Transformations of Sums and Products – Part 5
Question 6
\(6(i)\) Prove that,
\( \cos\theta\cos(60^{\circ}-\theta)\cos(60^{\circ}+\theta)
=\frac14\cos3\theta \)
Sol.
\(
\begin{aligned}
&\cos\theta\cos(60^{\circ}-\theta)\cos(60^{\circ}+\theta)\\
&=\frac12\cos\theta
[2\cos(60^{\circ}-\theta)\cos(60^{\circ}+\theta)]\\
&=\frac12\cos\theta
[\cos(60^{\circ}+\theta+60^{\circ}-\theta)
+\cos(60^{\circ}+\theta-60^{\circ}+\theta)]\\
&=\frac12\cos\theta[\cos120^{\circ}+\cos2\theta]\\
&=\frac12\cos\theta[\cos(90^{\circ}+30^{\circ})+\cos2\theta]\\
&=\frac12\cos\theta[-\sin30^{\circ}+\cos2\theta]\\
&=\frac12\cos\theta\times\left(-\frac12\right)
+\frac14\times2\cos2\theta\cos\theta\\
&=-\frac14\cos\theta
+\frac14[\cos(2\theta+\theta)+\cos(2\theta-\theta)]\\
&=-\frac14\cos\theta
+\frac14[\cos3\theta+\cos\theta]\\
&=-\frac14\cos\theta
+\frac14\cos3\theta+\frac14\cos\theta\\
&=\frac14\cos3\theta
\quad\text{(proved)}
\end{aligned}
\)
\(6(ii)\) Prove that,
\(4\sin\theta\sin\left(\frac{\pi}{3}+\theta\right)
\sin\left(\frac{\pi}{3}-\theta\right)=\sin3\theta\)
Sol.
\(
\begin{aligned}
&4\sin\theta
\sin\left(\frac{\pi}{3}+\theta\right)
\sin\left(\frac{\pi}{3}-\theta\right)\\
&=2\sin\theta\times
2\sin\left(\frac{\pi}{3}+\theta\right)
\sin\left(\frac{\pi}{3}-\theta\right)\\
&=2\sin\theta
\left[
\cos\left(\frac{\pi}{3}+\theta-\frac{\pi}{3}+\theta\right)
-\cos\left(\frac{\pi}{3}+\theta+\frac{\pi}{3}-\theta\right)
\right]\\
&=2\sin\theta
\left[\cos2\theta-\cos\left(\pi-\frac{\pi}{3}\right)\right]\\
&=2\sin\theta\cos2\theta
-2\sin\theta\left(-\cos\frac{\pi}{3}\right)\\
&=\sin(2\theta+\theta)-\sin(2\theta-\theta)
+2\sin\theta\times\frac12\\
&=\sin3\theta-\sin\theta+\sin\theta\\
&=\sin3\theta
\end{aligned}
\)
\(6(iii)\) Prove that,
\(\cos\alpha\cos(120^{\circ}+\alpha)
\cos(240^{\circ}+\alpha)=\frac14\cos3\alpha\)
Sol.
\(
\begin{aligned}
&\cos\alpha\cos(120^{\circ}+\alpha)
\cos(240^{\circ}+\alpha)\\
&=\frac12\cos\alpha
\times2\cos(120^{\circ}+\alpha)
\cos(240^{\circ}+\alpha)\\
&=\frac12\cos\alpha
[\cos(240^{\circ}+\alpha+120^{\circ}+\alpha)
+\cos(240^{\circ}+\alpha-120^{\circ}-\alpha)]\\
&=\frac12\cos\alpha
[\cos(360^{\circ}+2\alpha)+\cos120^{\circ}]\\
&=\frac12\cos\alpha
[\cos2\alpha+\cos(180^{\circ}-60^{\circ})]\\
&=\frac12\cos\alpha[\cos2\alpha-\cos60^{\circ}]\\
&=\frac14\times2\cos2\alpha\cos\alpha
-\frac12\cos\alpha\times\frac12\\
&=\frac14[\cos(2\alpha+\alpha)+\cos(2\alpha-\alpha)]
-\frac14\cos\alpha\\
&=\frac14[\cos3\alpha+\cos\alpha]
-\frac14\cos\alpha\\
&=\frac14\cos3\alpha
+\frac14\cos\alpha-\frac14\cos\alpha\\
&=\frac14\cos3\alpha
\quad\text{(proved)}
\end{aligned}
\)
Question 7 – Find the Value
\(7(i)\)
\(\sin78^{\circ}-\sin18^{\circ}
+\sin30^{\circ}-\sin42^{\circ}\)
\(
\begin{aligned}
&\sin78^{\circ}-\sin42^{\circ}
-\sin18^{\circ}+\sin30^{\circ}\\
&=2\cos\frac{78^{\circ}+42^{\circ}}2
\sin\frac{78^{\circ}-42^{\circ}}2
-\sin18^{\circ}+\frac12\\
&=2\cos60^{\circ}\sin18^{\circ}
-\sin18^{\circ}+\frac12\\
&=2\times\frac12\sin18^{\circ}
-\sin18^{\circ}+\frac12\\
&=\frac12
\end{aligned}
\)
\(7(ii)\)
\(\sec20^{\circ}\sec40^{\circ}\sec80^{\circ}\)
\(
\begin{aligned}
&\sec20^{\circ}\sec40^{\circ}\sec80^{\circ}\\
&=\frac{2}
{(2\cos20^{\circ}\cos40^{\circ})\cos80^{\circ}}\\
&=\frac{2}
{(\cos60^{\circ}+\cos20^{\circ})\cos80^{\circ}}\\
&=\frac{2}
{(\frac12+\cos20^{\circ})\cos80^{\circ}}\\
&=\frac{2}
{\frac12\cos80^{\circ}
+\frac12\times2\cos80^{\circ}\cos20^{\circ}}\\
&=\frac{4}
{\cos80^{\circ}
+(\cos100^{\circ}+\cos60^{\circ})}\\
&=\frac{4}
{\sin(90^{\circ}-80^{\circ})
+\cos(90^{\circ}+10^{\circ})
+\frac12}\\
&=\frac{4}
{\sin10^{\circ}-\sin10^{\circ}+\frac12}\\
&=8\quad\text{(ans.)}
\end{aligned}
\)
\(7(iii)\)
\(\sqrt3\cot20^{\circ}\cot40^{\circ}\cot80^{\circ}\)
\(
\begin{aligned}
&\sqrt3\cot20^{\circ}\cot40^{\circ}\cot80^{\circ}\\
&=\sqrt3\times
\frac{\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}
{\sin20^{\circ}\sin40^{\circ}\sin80^{\circ}}\\
&=\sqrt3\times
\frac{(2\cos20^{\circ}\cos40^{\circ})\cos80^{\circ}}
{(2\sin20^{\circ}\sin40^{\circ})\sin80^{\circ}}\\
&=\sqrt3\times
\frac{(\cos60^{\circ}+\cos20^{\circ})\cos80^{\circ}}
{(\cos20^{\circ}-\cos60^{\circ})\sin80^{\circ}}\\
&=\sqrt3\times
\frac{\frac12\cos80^{\circ}
+\frac12(2\cos80^{\circ}\cos20^{\circ})}
{\frac12(2\sin80^{\circ}\cos20^{\circ})
-\frac12\sin80^{\circ}}\\
&=\sqrt3\times
\frac{\cos80^{\circ}
+\cos(80^{\circ}+20^{\circ})
+\cos(80^{\circ}-20^{\circ})}
{\sin(80^{\circ}+20^{\circ})
+\sin(80^{\circ}-20^{\circ})
-\sin80^{\circ}}\\
&=\sqrt3\times
\frac{\cos80^{\circ}+\cos100^{\circ}+\cos60^{\circ}}
{\sin100^{\circ}+\sin60^{\circ}-\sin80^{\circ}}\\
&=\sqrt3\times
\frac{\sin(90^{\circ}-80^{\circ})
+\cos(90^{\circ}+10^{\circ})
+\cos60^{\circ}}
{\sin(90^{\circ}+10^{\circ})
+\sin60^{\circ}
-\sin(90^{\circ}-10^{\circ})}\\
&=\sqrt3\times
\frac{\sin10^{\circ}-\sin10^{\circ}+\cos60^{\circ}}
{\cos10^{\circ}+\sin60^{\circ}-\cos10^{\circ}}\\
&=\sqrt3\times
\frac{\cos60^{\circ}}{\sin60^{\circ}}\\
&=\sqrt3\times\cot60^{\circ}\\
&=\sqrt3\times\frac1{\sqrt3}\\
&=1
\end{aligned}
\)
\(7(iv)\)
\(\frac{1}{\sin10^{\circ}}-2\sin70^{\circ}\)
\(
\begin{aligned}
&\frac{1}{\sin10^{\circ}}-2\sin70^{\circ}\\
&=\frac{1-2\times2\sin70^{\circ}\sin10^{\circ}}
{2\sin10^{\circ}}\\
&=\frac{1-2[
\cos(70^{\circ}-10^{\circ})
-\cos(70^{\circ}+10^{\circ})]}
{2\sin10^{\circ}}\\
&=\frac{1-2(\cos60^{\circ}-\cos80^{\circ})}
{2\sin10^{\circ}}\\
&=\frac{1-2(\frac12-\sin10^{\circ})}
{2\sin10^{\circ}}\\
&=\frac{1-1+2\sin10^{\circ}}
{2\sin10^{\circ}}\\
&=1
\end{aligned}
\)
Question 8 – Prove the Following Identities
\(8(i)\)
\(4\sin23^{\circ}\sin37^{\circ}\sin83^{\circ}
=\cos21^{\circ}\)
Sol.
\(
\begin{aligned}
&4\sin23^{\circ}\sin37^{\circ}\sin83^{\circ}\\
&=2\sin23^{\circ}
(2\sin37^{\circ}\sin83^{\circ})\\
&=2\sin23^{\circ}
[\cos(83^{\circ}-37^{\circ})
-\cos(83^{\circ}+37^{\circ})]\\
&=2\sin23^{\circ}
[\cos46^{\circ}-\cos120^{\circ}]\\
&=2\sin23^{\circ}\cos46^{\circ}
-2\sin23^{\circ}\cos(90^{\circ}+30^{\circ})\\
&=2\cos46^{\circ}\sin23^{\circ}
+2\sin23^{\circ}\sin30^{\circ}\\
&=\sin(46^{\circ}+23^{\circ})
-\sin(46^{\circ}-23^{\circ})
+2\sin23^{\circ}\times\frac12\\
&=\sin69^{\circ}-\sin23^{\circ}+\sin23^{\circ}\\
&=\sin69^{\circ}\\
&=\cos(90^{\circ}-69^{\circ})\\
&=\cos21^{\circ}
\quad\text{(proved)}
\end{aligned}
\)
\(8(ii)\)
\(\cos2A+\cos4A+\cos6A+\cos8A
=4\cos A\cos2A\cos5A\)
Sol.
\(
\begin{aligned}
&\cos2A+\cos4A+\cos6A+\cos8A\\
&=(\cos2A+\cos6A)+(\cos4A+\cos8A)\\
&=2\cos\left(\frac{6A+2A}{2}\right)
\cos\left(\frac{6A-2A}{2}\right)\\
&\quad+2\cos\left(\frac{8A+4A}{2}\right)
\cos\left(\frac{8A-4A}{2}\right)\\
&=2\cos4A\cos2A+2\cos6A\cos2A\\
&=2\cos2A(\cos4A+\cos6A)\\
&=2\cos2A\times2\cos\left(\frac{6A+4A}{2}\right)
\cos\left(\frac{6A-4A}{2}\right)\\
&=4\cos2A\cos5A\cos A\\
&=4\cos A\cos2A\cos5A
\quad\text{(proved)}
\end{aligned}
\)
\(8(iii)\)
\(\frac{\sin(A+B)-2\sin A+\sin(A-B)}
{\cos(A+B)-2\cos A+\cos(A-B)}
=\tan A\)
Sol.
\(
\begin{aligned}
&\frac{\sin(A+B)-2\sin A+\sin(A-B)}
{\cos(A+B)-2\cos A+\cos(A-B)}\\
&=\frac{[\sin(A+B)+\sin(A-B)]-2\sin A}
{[\cos(A+B)+\cos(A-B)]-2\cos A}\\
&=\frac{2\sin A\cos B-2\sin A}
{2\cos A\cos B-2\cos A}\\
&=\frac{2\sin A(\cos B-1)}
{2\cos A(\cos B-1)}\\
&=\tan A
\quad\text{(proved)}
\end{aligned}
\)
\(8(iv)\)
\(\frac{\sin\theta+\sin3\theta+\sin5\theta+\sin7\theta}
{\cos\theta+\cos3\theta+\cos5\theta+\cos7\theta}
=\tan4\theta\)
Sol.
\(
\begin{aligned}
&\frac{\sin\theta+\sin3\theta+\sin5\theta+\sin7\theta}
{\cos\theta+\cos3\theta+\cos5\theta+\cos7\theta}\\
&=\frac{(\sin\theta+\sin5\theta)
+(\sin3\theta+\sin7\theta)}
{(\cos\theta+\cos5\theta)
+(\cos3\theta+\cos7\theta)}\\
&=\frac{2\sin3\theta\cos2\theta
+2\sin5\theta\cos2\theta}
{2\cos3\theta\cos2\theta
+2\cos5\theta\cos2\theta}\\
&=\frac{2\cos2\theta
(\sin3\theta+\sin5\theta)}
{2\cos2\theta
(\cos3\theta+\cos5\theta)}\\
&=\frac{2\sin4\theta\cos\theta}
{2\cos4\theta\cos\theta}\\
&=\tan4\theta
\quad\text{(proved)}
\end{aligned}
\)
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