NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

Question:1

Prove that
$\frac{tanA+secA-1}{tanA-secA+1}=\frac{1+sinA}{cosA}$

Answer:

$$\begin{gathered} L.H.S=\frac{tanA+secA-1}{tanA-secA+1} \\ =\frac{tanA+secA-({\sec }^{2}A-{\tan }^{2}A)}{tanA-secA+1} \\ =\frac{tanA+secA-\left[(secA+tanA)(secA-tanA)\right]}{tanA-secA+1} \\ =\frac{(secA+tanA)[1-(secA-tanA)]}{tanA-secA+1} \end{gathered}$$

$$\begin{gathered} =\frac{(secA+tanA)[1-secA+tanA]}{tanA-secA+1} \\ =secA+tanA \\ =\frac{1}{cosA}+\frac{sinA}{cosA} \\ =\frac{1+sinA}{cosA}=R.H.S \end{gathered}$$

Question:2

If $[2\sin \alpha /(1+\cos \alpha +\sin \alpha )]=y$, then prove that $(1-\cos \alpha +\sin \alpha )/(1+\sin \alpha )$ is also equal to y.

$[$ Hint: Express $\frac{1-\cos \alpha +\sin \alpha }{1+\sin \alpha }=\frac{1-\cos \alpha +\sin \alpha }{1+\sin \alpha }\cdot \frac{1+\cos \alpha +\sin \alpha }{1+\cos \alpha +\sin \alpha }]$

Answer:

$$\begin{gathered} y=\frac{2sin\alpha }{1+cos\alpha +sin\alpha }=\frac{2sin\alpha }{1+cos\alpha +sin\alpha }\ast \frac{1+sin\alpha -cos\alpha }{1+sin\alpha -cos\alpha } \\ =\frac{2sin\alpha (1-cos\alpha +sin\alpha )}{{(1+sin\alpha )}^2-{\cos }^2\alpha } \\ =\frac{2sin\alpha (1-cos\alpha +sin\alpha )}{1+{\sin }^2\alpha +2sin\alpha -{\cos }^2\alpha } \end{gathered}$$
$$\begin{gathered} =\frac{2sin\alpha (1-cos\alpha +sin\alpha )}{(1-{\cos }^2\alpha )+{\sin }^2\alpha +2sin\alpha } \\ =\frac{2sin\alpha (1-cos\alpha +sin\alpha )}{2si{n}^2\alpha +2sin\alpha } \\ =\frac{2sin\alpha (1-cos\alpha +sin\alpha )}{2sin\alpha (1+sin\alpha )}=\frac{(1-cos\alpha +sin\alpha )}{(1+sin\alpha )}=y \end{gathered}$$

Question:3

If $msin\theta =nsin(\theta +2\alpha )$, then prove that
$tan(\theta +\alpha )cot\alpha =(m+n)/(m-n)$
[Hints: Express $sin(\theta +2\alpha )/sin\theta =m/n$ and apply componendo and dividendo]

Answer:

$$\begin{gathered} Giventhatmsin\theta =nsin(\theta +2\alpha ) \\ \frac{\sin (\theta +2\alpha )}{sin\theta }=\frac{m}{n} \\ \frac{\sin (\theta +2\alpha )+sin\theta }{\sin (\theta +2\alpha )-sin\theta }=\frac{m+n}{m-n} \end{gathered}$$

$\frac{2\sin \left(\frac{\theta +2\alpha +\theta }{2}\right)\cdot \cos \left(\frac{\theta +2\alpha -\theta }{2}\right)}{2\cos \left(\frac{\theta +2\alpha +\theta }{2}\right)\cdot \sin \left(\frac{\theta +2\alpha -\theta }{2}\right)}=\frac{m+n}{m-n}$

$\tan (\theta +\alpha ).cot\alpha =\frac{m+n}{m-n}$

Question:4

If $\cos (\alpha +\beta )=\frac{4}{5}\text{ and }\sin (\alpha -\beta )=\frac{5}{13}$ where α lie between 0 and π/4, find value of tan 2α
[Hint: Express $tan2\alpha astan(\alpha +\beta +\alpha -\beta ]$

Answer:

$$\begin{gathered} \cos (\alpha +\beta )=\frac{4}{5}\text{~ so,}\tan (\alpha +\beta )=\frac{3}{4} \\ \text{~~~~ And}\sin (\alpha -\beta )=\frac{5}{13}\tan (\alpha -\beta )=\frac{5}{12} \\ tan2\alpha =\tan [\alpha +\beta +\alpha -\beta ] \end{gathered}$$
$$\begin{gathered} =\tan [(\alpha +\beta )+(\alpha -\beta )] \\ =\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{1-\tan (\alpha +\beta )\tan (\alpha -\beta )} \end{gathered}$$

$=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\ast \frac{5}{12}}=\frac{56}{33}$

Question:5

If $tanx=b/a$ then find the value of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$

Answer:

$$\begin{gathered} tanx=\frac{b}{a} \\ \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{a+b+a-b}{\sqrt{(a-b)(a+b)}} \\ =\frac{2a}{\sqrt{a^2-b^2}} \\ =\frac{2a}{a\sqrt{1-\frac{b^2}{a^2}}} \end{gathered}$$

$=\frac{2}{\sqrt{1-{\tan }^{2}x}}=\frac{2}{\sqrt{1-({\sin }^{2}x)/({\cos }^{2}x)}}=2cosx/\left(\sqrt{cos2x}\right)$

Question:6

Prove that $\cos \theta \cos \frac{\theta }{2}-\cos 3\theta \cos \frac{9\theta }{2}=\sin 7\theta \sin 4\theta$
$[\text{Hint:Express L.H.S. }=\frac{1}{2}[2cos\theta cos\theta /2-2cos3\theta cos9\theta /2]$

Answer:

$L.H.S=cos\theta \cos \left(\frac{\theta }{2}\right)-cos3\theta \cos \left(\frac{9\theta }{2}\right)$

$=\frac{1}{2}[2cos\theta \cos \left(\frac{\theta }{2}\right)]-\frac{1}{2}[2cos3\theta \cos \left(\frac{9\theta }{2}\right)]$

$=\frac{1}{2}[\cos (\theta +\frac{\theta }{2})+\cos (\theta -\frac{\theta }{2})]-\frac{1}{2}[\cos (3\theta +\frac{9\theta }{2})+\cos (3\theta -\frac{9\theta }{2})]$

$=\frac{1}{2}[\cos \frac{3\theta }{2}+\cos \frac{\theta }{2}-\cos \frac{15\theta }{2}-\cos (-\frac{3\theta }{2})]$

$=\frac{1}{2}[\cos \frac{3\theta }{2}+\cos \frac{\theta }{2}-\cos \frac{15\theta }{2}-\cos \frac{3\theta }{2}]$

$$\begin{gathered} =\frac{1}{2}[\cos \frac{\theta }{2}-\cos \frac{15\theta }{2}] \\ =\frac{1}{2}[-2\sin \left(\frac{\frac{\theta }{2}+\frac{15\theta }{2}}{2}\right)\sin \left(\frac{\frac{\theta }{2}-\frac{15\theta }{2}}{2}\right)] \end{gathered}$$

$=\frac{1}{2}[-2\sin 4\theta \sin (-\frac{7\theta }{2})]$
$=-\sin \left(4\theta \right)\sin (-\frac{7\theta }{2})=\sin \left(4\theta \right)\sin \left(\frac{7\theta }{2}\right)$

Question:7

If $a\cos \theta +b\sin \theta =manda\sin \theta -bcos\theta =n$, then show that $m^2+n^2=a^2+b^2$

Answer:

$acos\theta +bsin\theta =masin\theta -bcos\theta =n$

$R.H.S=m^2+n^2={(acos\theta +bsin\theta )}^2+{(asin\theta -bcos\theta )}^2$

$=a^2{\cos }^2\theta +b^2{\sin }^2\theta +2absin\theta cos\theta +a^2{\sin }^2\theta +b^2{\cos }^2\theta -2absin\theta cos\theta$

$$\begin{gathered} =a^2({\cos }^2\theta +{\sin }^2\theta )+b^2({\cos }^2\theta +{\sin }^2\theta ) \\ =a^2+b^2 \end{gathered}$$

Question:8

Find the value of $tan{22}^{\circ }{30}^{{}^{\prime }}$.
$\begin{array}{rl}& \text{ [Hint: Let }\theta ={45}^{\circ }\text{ , use }\\ & \tan \frac{\theta }{2}=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}=\frac{\sin \theta }{1+\cos \theta }\end{array}]$

Answer:

$Let{22}^0{30}^{{}^{\prime }}=\frac{\theta }{2}\tan {22}^0{30}^{{}^{\prime }}=\tan \frac{\theta }{2}$

$=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}$

$=\frac{sin\theta }{1+cos\theta }$

$$\begin{gathered} Putting\theta =45 \\ \frac{sin\theta }{1+cos\theta }=\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} \\ =\frac{1}{\sqrt{2}+1}=(\sqrt{2}-1)\left[Onrationalising\right] \end{gathered}$$

Question:9

Prove that $\sin 4A=4\sin A{\cos }^{3}A-4\cos A{\sin }^{3}A$.

Answer:

$L.H.S\sin 4A=\sin (A+3A)=sinA\cos 3A+\cos A\sin 3A$

$=\sin A(4{\cos }^{3}A-3\cos A)+\cos A(3sinA-4{\sin }^{3}A)$

$=4\sin A{\cos }^{3}A-3sinA\cos A+3sinA\cos A-4\cos A{\sin }^{3}A$

$=4sinA{\cos }^{3}A-4cosA{\sin }^{3}A=R.H.S$

Question:10

If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$, then prove that

$m^2-n^2=4\sin \theta \tan \theta$

$[$ Hint: $m+n=2\tan \theta ,m-n=2\sin \theta$, then use $[m^2-n^2]=(m+n)(m-n)]$

Answer:

$\tan \theta +\sin \theta =mand\tan \theta -\sin \theta =n$$L.H.S=m^2-n^2=(m+n)(m-n)$$=[(\tan \theta +\sin \theta )+(\tan \theta -\sin \theta )].[(\tan \theta +\sin \theta )-(\tan \theta -\sin \theta )]$$$\begin{gathered} =( \\ tan\theta +\sin \theta +\tan \theta -\sin \theta ).( \\ tan\theta +\sin \theta -\tan \theta +\sin \theta ) \end{gathered}$$$=2\tan \theta .2\sin \theta =4\sin \theta \tan \theta =R.H.S$

Question:11

If $tan(A+B)=p,tan(A-B)=q$, then show that $tan2A=(p+q)/(1-pq)$.
[Hint: Use $2A=(A+B)+(A-B)$]

Answer:

$$\begin{gathered} \tan (A+B)=p,tan(A-B)=q \\ tan2A=tan(A+B+A-B)=tan[(A+B)+(A-B)] \\ =\frac{ \\ tan(A+B)+ \\ tan(A-B)}{1- \\ tan(A+B). \\ tan(A-B)} \\ =\frac{p+q}{1-pq} \end{gathered}$$

Question:12

If $\cos \alpha +\cos \beta =0=\sin \alpha +\sin \beta$, then prove that $\cos 2\alpha +\cos 2\beta =-2\cos (\alpha +\beta )$. [Hint: $(\cos \alpha +\cos \beta {)}^2-(\sin \alpha +\sin \beta {)}^2=0]$

Answer:

Given that: $\cos \alpha +\cos \beta =0and\sin \alpha +\sin \beta =0$

$So,{(\cos \alpha +\cos \beta )}^2-{(\sin \alpha +\sin \beta )}^2=0$

$$\begin{gathered} ( \\ co{s}^2\alpha + \\ co{s}^2\beta +2\cos \alpha \cos \beta )-({\sin }^2\alpha + \\ si{n}^2\beta +2\sin \alpha \sin \beta )=0 \end{gathered}$$

$$\begin{gathered} ( \\ co{s}^2\alpha - \\ si{n}^2\alpha )+( \\ co{s}^2\beta - \\ si{n}^2\beta )+2(\cos \alpha \cos \beta -\sin \alpha \sin \beta )=0 \end{gathered}$$

$cos2\alpha +cos2\beta +2\cos (\alpha +\beta )=0$

$cos2\alpha +\cos 2\beta =-2\cos (\alpha +\beta )$

Question:13

If $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$ , then show that $\frac{\tan x}{\tan y}=\frac{a}{b}$ [Hint: Use componendo and Dividendo]

Answer:

$$\begin{gathered} \frac{\sin (x+y)}{ \\ sin(x-y)}=\frac{a+b}{a-b} \\ \frac{ \\ sin(x+y)+ \\ sin(x-y)}{ \\ sin(x-y)- \\ sin(x-y)}=\frac{a+b+a-b}{a+b-a+b} \end{gathered}$$

$\frac{2\sin \frac{x+y+x-y}{2}\cos \frac{x+y-x+y}{2}}{2\cos \frac{x+y+x-y}{2}\sin \frac{x+y-x+y}{2}}=\frac{2a}{2b}$

$\frac{\sin x\cos y}{\cos x\sin y}=\frac{a}{b}$

$\frac{\tan x}{\tan y}=\frac{a}{b}$

Question:14

If $\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$ then show that $\sin \alpha +\cos \alpha =\surd 2\cos \theta$
[Hint: Express $\tan \theta =\tan (\alpha -\pi /4)$, $\theta =\alpha -\pi /4$]

Answer:

$$\begin{gathered} \tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha } \\ \tan \theta =\frac{\frac{\sin \alpha -\cos \alpha }{\cos \alpha }}{\frac{\sin \alpha +\cos \alpha }{\cos \alpha }}=\frac{\tan \alpha -1}{\tan \alpha +1} \\ =(\tan \alpha - \\ tan\frac{\pi }{4})/(1+ \\ tan\frac{\pi }{4}\tan \alpha ) \\ \tan \theta =tan(\alpha -\frac{\pi }{4}) \end{gathered}$$

$\theta =\alpha -\frac{\pi }{4}$

$\cos \theta =\cos (\alpha -\frac{\pi }{4})$

$$\begin{gathered} \cos \theta =\cos \alpha cos\frac{\pi }{4}+\sin \alpha sin\frac{\pi }{4} \\ \cos \theta =\cos \alpha .\frac{1}{\sqrt{2}}+\sin \alpha .\frac{1}{\sqrt{2}} \end{gathered}$$

$\sqrt{2}\cos \theta =\cos \alpha +\sin \alpha$

Hence, proved.

Question:15

If $\sin \theta +\cos \theta =1$ , then find the general value of $\theta$

Answer:

Given that $\sin \theta +\cos \theta =1\text{Dividing both sides by}\sqrt{1^2+1^2}=\sqrt{2}$

$$\begin{gathered} \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{1}{\sqrt{2}} \\ cos(\theta -\frac{\pi }{4})=\cos \frac{\pi }{4} \\ \theta -\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4},nϵZ \end{gathered}$$

$$\begin{gathered} \theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{4} \\ \theta =2n\pi +\frac{\pi }{4}+\frac{\pi }{4}or\theta =2n\pi -\frac{\pi }{4}+\frac{\pi }{4} \\ \theta =2n\pi +\frac{\pi }{2}or\theta =2n\pi ,nϵZ \end{gathered}$$

Question:16

Find the most general value of $\theta$ satisfying the equation $\tan \theta =-1and\cos \theta =1/\sqrt{2}$

Answer:

Given that

$$\begin{gathered} \tan \theta =-1and\cos \theta =\frac{1}{\sqrt{2}} \\ \cos \theta \text{ is positive and }\tan \theta \text{is negative in fourth quadrant}\tan \theta =-1 \\ \tan \theta =\tan (-\frac{\pi }{4}) \\ \tan \theta =\tan (2\pi -\frac{\pi }{4}) \end{gathered}$$

$$\begin{gathered} \tan \theta =\tan \frac{7\pi }{4} \\ \theta =\frac{7\pi }{4} \\ \cos \theta =\frac{1}{\sqrt{2}} \\ \cos \theta =\cos \frac{\pi }{4} \end{gathered}$$

$$\begin{gathered} \cos \theta =\cos (2\pi -\frac{\pi }{4}) \\ \cos \theta =\cos \frac{7\pi }{4} \\ \theta =\frac{7\pi }{4} \\ \text{So, general solution is }\theta =2n\pi +\frac{7\pi }{4} \end{gathered}$$

Question:17

If $\cot \theta +\tan \theta =2cosec\theta$, then find the general value of $\theta$.

Answer:

$\text{Given that}\cot \theta +tan\theta =2cosec\theta$

$$\begin{gathered} \frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }=\frac{2}{\sin \theta } \\ \frac{ \\ si{n}^2\theta + \\ co{s}^2\theta }{\sin \theta \cos \theta }=\frac{2}{\sin \theta } \\ \frac{1}{\sin \theta \cos \theta }=\frac{2}{\sin \theta } \\ 2\sin \theta \cos \theta =\sin \theta \end{gathered}$$
$$\begin{gathered} \sin \theta (2\cos \theta -1)=0 \\ \sin \theta =0or2\cos \theta -1=0or\cos \theta =\frac{1}{2} \\ Now,\sin \theta =0\Rightarrow \theta =n\pi ,nϵZ \\ \cos \theta =\frac{1}{2}\Rightarrow \cos \theta =\cos \frac{\pi }{3} \\ \theta =2n\pi \pm \frac{\pi }{3} \end{gathered}$$
Hence, general values of $\theta$ is $2n\pi \pm \frac{\pi }{3}andn\pi ,nϵZ$

Question:18

If $2{\sin }^2\theta =3\cos \theta$, where $0\le \theta \le 2\pi$, then find the value of $\theta$.

Answer:

$$\begin{gathered} \text{Given that }2{\sin }^2\theta =3\cos \theta \\ 2(1-{\cos }^2\theta )=3\cos \theta \\ 2-2{\cos }^2\theta -3\cos \theta =0 \\ 2{\cos }^2\theta +3\cos \theta -2=0 \end{gathered}$$

$(\cos \theta +2)(2\cos \theta -1)=0\left[\text{On factorising}\right]$

$\cos \theta +2=0or2\cos \theta -1=0$

$cos\theta \ne -2$

$So,2\cos \theta -1=0,\cos \theta =\frac{1}{2}$
$$\begin{gathered} or\theta =\frac{\pi }{3}or2\pi -\frac{\pi }{3} \\ \theta =\frac{\pi }{3}or\frac{5\pi }{3} \end{gathered}$$

Question:19

If $secx\cos 5x+1=0,where0<x\le \pi /2$, then find the value of x.

Answer:

$$\begin{gathered} \text{Given that }\sec x\cos 5x+1=0 \\ \frac{1}{\cos x}.\cos 5x+1=0 \\ cos5x+\cos x=0 \\ \text{~ 2}\cos \left(\frac{5x+x}{2}\right)\cos \left(\frac{5x-x}{2}\right)=0 \end{gathered}$$
$$\begin{gathered} \cos 3x.\cos 2x=0 \\ \cos 3x=0or\cos 2x=0 \\ 3x=\frac{\pi }{2}or2x=\frac{\pi }{2} \\ x=\frac{\pi }{6}orx=\frac{\pi }{4} \end{gathered}$$

Question:20

If $\sin (\theta +\alpha )=a$ and $\sin (\theta +\beta )=b$, then prove that $\cos 2(\alpha -\beta )-4ab\cos (\alpha -\beta )=1-2a^2-2b^2$

Answer:

$$\begin{gathered} \text{Given that }\sin (\theta +\alpha )=aand\sin (\theta +\beta )=b\dots \dots \dots \dots \left(i\right) \\ \cos (\alpha -\beta )=\cos [\theta +\alpha -\theta -\beta ] \\ =\cos [(\theta +\alpha )-(\theta +\beta )] \end{gathered}$$

$=\cos (\theta +\alpha )\cos (\theta +\beta )+\sin (\theta +\alpha )\sin (\theta +\beta )$$=\sqrt{1-{\sin }^2(\theta +\alpha )}.\sqrt{1-{\sin }^2(\theta +\beta )}+\sin (\theta +\alpha )\sin (\theta +\beta )$$=\sqrt{(1-a^2)(1-b^2)}+ab$$\cos (\alpha -\beta )=ab+\sqrt{1-a^2-b^2+a^2{b}^2}$

$$\begin{gathered} Now,\cos 2(\alpha -\beta )-4ab\cos (\alpha -\beta )=2{\cos }^2(\alpha -\beta )-1-4ab\cos (\alpha -\beta ) \\ =2{[ab+\sqrt{1-a^2-b^2+a^2{b}^2}]}^2-1-4ab[ab+\sqrt{1-a^2-b^2+a^2{b}^2}] \end{gathered}$$

$$\begin{gathered} =2[a^2{b}^2+1-a^2-b^2+a^2{b}^2+2ab\sqrt{1-a^2-b^2+a^2{b}^2}]-1-4a^2{b}^2-4ab\sqrt{1-a^2-b^2+a^2{b}^2} \\ =1-2a^2-2b^2 \end{gathered}$$

Question:21

If $\cos (\theta +\varphi )=m\cos (\theta -\varphi )$, then prove that $\tan \theta =((1-m)/(1+m))cot\varphi$
[Hint: Express $\cos (\theta +\varphi )/\cos (\theta -\varphi )=m/1$ and apply Componendo and Dividendo]

Answer:

$$\begin{gathered} \text{Given that}\cos (\theta +\varphi )=m\cos (\theta -\varphi ) \\ \frac{\cos (\varphi +\theta )}{\cos (\theta -\varphi )}=\frac{m}{1} \end{gathered}$$

$\text{~~Using~componendo~and dividend rule, }\frac{\cos (\varphi +\theta )+\cos (\theta -\varphi )}{\cos (\varphi +\theta )-\cos (\theta -\varphi )}=\frac{m+1}{m-1}$
$$\begin{gathered} \frac{\cos \theta \cos \varphi }{-\sin \theta \sin \varphi }=\frac{m+1}{m-1} \\ -cot\theta cot\varphi =\frac{m+1}{m-1} \\ \frac{-cot\varphi }{\tan \theta }=\frac{m+1}{m-1} \\ \tan \theta (1+m)=(1-m)cot\varphi \end{gathered}$$

$\tan \theta =\frac{1-m}{1+m}cot\varphi$

Question:22

Find the value of the expression
$=3[{\sin }^4(\frac{3\pi }{2}-\alpha )+{\sin }^4(3\pi +\alpha )]-2[{\sin }^6(\frac{\pi }{2}+\alpha )+{\sin }^6(5\pi -\alpha )]$

Answer:

$=3[{\sin }^4(\frac{3\pi }{2}-\alpha )+{\sin }^4(3\pi +\alpha )]-2[{\sin }^6(\frac{\pi }{2}+\alpha )+{\sin }^6(5\pi -\alpha )]$$=3[{\cos }^4\alpha +{\sin }^4(\pi +\alpha )]-2[{\cos }^6\alpha +{\sin }^6(\pi -\alpha )]$$=3[{\cos }^4\alpha +{\sin }^4\alpha ]-2[{\cos }^6\alpha +{\sin }^6\alpha ]$$=3[{\cos }^4\alpha +{\sin }^4\alpha +2{\sin }^2\alpha {\cos }^2\alpha -2{\sin }^2\alpha {\cos }^2\alpha ]-2[{({\cos }^2\alpha +{\sin }^2\alpha )}^3-3{\cos }^2\alpha {\sin }^2\alpha ({\cos }^2\alpha +{\sin }^2\alpha )]$$$