Question

Prove that sinA + sin3A + sin5A + sin7A / cosA + cos3A + cos5A + cos7A = tan4A​

Answer 1

Answer:

SinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A

SinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)

SinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)=(2sin4Acos3A+2sin4AcosA)/(2cos4Acos3A+2cos4AcosA)

SinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)=(2sin4Acos3A+2sin4AcosA)/(2cos4Acos3A+2cos4AcosA)=2sin4A(cos3A+cosA)/2cos4A(cos3A+cosA)

SinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)=(2sin4Acos3A+2sin4AcosA)/(2cos4Acos3A+2cos4AcosA)=2sin4A(cos3A+cosA)/2cos4A(cos3A+cosA)=sin4A/cos4A

SinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)=(2sin4Acos3A+2sin4AcosA)/(2cos4Acos3A+2cos4AcosA)=2sin4A(cos3A+cosA)/2cos4A(cos3A+cosA)=sin4A/cos4A=tan4A (Proved)

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Answer 2

We know the sum - to - product identities,

• $\sf{\sin A+\sin B=2\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)}$

• $\sf{\cos A+\cos B=2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)}$

Thus,

$\longrightarrow\sf{\dfrac{\sin A+\sin(3A)+\sin(5A)+\sin(7A)}{\cos A+\cos(3A)+\cos(5A)+\cos(7A)}=\dfrac{\sin A+\sin(7A)+\sin(3A)+\sin(5A)}{\cos A+\cos(7A)+\cos(3A)+\cos(5A)}}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{\sin A+\sin(3A)+\sin(5A)+\sin(7A)}{\cos A+\cos(3A)+\cos(5A)+\cos(7A)}}\\\\=\ \ &\sf{\dfrac{2\sin\left(\dfrac{A+7A}{2}\right)\cos\left(\dfrac{A-7A}{2}\right)+2\sin\left(\dfrac{3A+5A}{2}\right)\cos\left(\dfrac{3A-5A}{2}\right)}{2\cos\left(\dfrac{A+7A}{2}\right)\cos\left(\dfrac{A-7A}{2}\right)+2\cos\left(\dfrac{3A+5A}{2}\right)\cos\left(\dfrac{3A-5A}{2}\right)}}\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{\sin A+\sin(3A)+\sin(5A)+\sin(7A)}{\cos A+\cos(3A)+\cos(5A)+\cos(7A)}}\\\\=\ \ &\sf{\dfrac{2\sin\left(\dfrac{8A}{2}\right)\cos\left(\dfrac{-6A}{2}\right)+2\sin\left(\dfrac{8A}{2}\right)\cos\left(\dfrac{-2A}{2}\right)}{2\cos\left(\dfrac{8A}{2}\right)\cos\left(\dfrac{-6A}{2}\right)+2\cos\left(\dfrac{8A}{2}\right)\cos\left(\dfrac{-2A}{2}\right)}}\end{aligned}$

Since $\sf{\cos(-x)=\cos x,}$

$\longrightarrow\sf{\dfrac{\sin A+\sin(3A)+\sin(5A)+\sin(7A)}{\cos A+\cos(3A)+\cos(5A)+\cos(7A)}=\dfrac{\sin\left(4A\right)\cos\left(3A\right)+\sin\left(4A\right)\cos A}{\cos\left(4A\right)\cos\left(3A\right)+\cos\left(4A\right)\cos A}}$

$\longrightarrow\sf{\dfrac{\sin A+\sin(3A)+\sin(5A)+\sin(7A)}{\cos A+\cos(3A)+\cos(5A)+\cos(7A)}=\dfrac{\sin\left(4A\right)[\cos\left(3A\right)+\cos A]}{\cos\left(4A\right)[\cos\left(3A\right)+\cos A]}}$

$\longrightarrow\sf{\dfrac{\sin A+\sin(3A)+\sin(5A)+\sin(7A)}{\cos A+\cos(3A)+\cos(5A)+\cos(7A)}=\dfrac{\sin\left(4A\right)}{\cos\left(4A\right)}}$

$\longrightarrow\underline{\underline{\sf{\dfrac{\sin A+\sin(3A)+\sin(5A)+\sin(7A)}{\cos A+\cos(3A)+\cos(5A)+\cos(7A)}=\tan(4A)}}}$

Hence Proved!