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Trigonometric Identities — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsTrigonometric Identities4 Marks
Question
Prove the following trigonometric identities.
If $\cos\text{A}+\cos^2\text{A}=1,$ prove that $\sin^{12}\theta+\sin^{10}\theta+3\sin^8\theta+\sin^6\theta+2\sin^4\theta+2\sin^2\theta-2=1.$
✓

Answer

Given,
$\cos\theta+\cos^2\theta=1$
We have to prove $\sin^{12}\theta+\sin^{10}\theta+3\sin^8\theta+\sin^6\theta+2\sin^4\theta+2\sin^2\theta-2=1.$
From the given equation, we have
$\cos\theta+\cos^2\theta=1$
$\Rightarrow \cos\theta=1-\cos^2\theta$
$\Rightarrow \cos\theta=\sin^2\theta$
$\Rightarrow \sin^2\theta=\cos\theta$
Therefore, we have
$\text{L.H.S}=\sin^{12}\theta+\sin^{10}\theta+3\sin^8\theta+\sin^6\theta+2\sin^4\theta+2\sin^2\theta-2$
$=(\sin^{12}\theta+\sin^{10}\theta+3\sin^8\theta+\sin^6\theta)+2(\sin^4\theta+\sin^2\theta)-2$
$=\{(\sin^4\theta)^3+3(\sin^4\theta)^2\sin^2\theta+3\sin^4\theta(\sin^2\theta)^2+(\sin^2\theta)^3\}$
$=2(\sin^4+\sin^2\theta)-2$
$=(\sin^4\theta+\sin^2\theta)^3+2(\sin^4\theta+\sin^2\theta)-2$
$=(\cos^2\theta+\cos\theta)^3+2(\cos^2\theta+\cos\theta)-2$
$=(1)^3+2(1)-2$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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