Question

Prove that:
$\dfrac{\sin A+\sin 3A+\sin 5A+\sin 7A}{\cos A+\cos 3A+\cos 5A+\cos 7A}=\tan 4A$

Answer 1

To solve the given trigonometric question, we should know some of the trigonometric properties, these are given below, $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$. We should also know the similar property for cosines, $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$. We should also know that $\cos (-x)=\cos x$. Using these properties, we will prove the given statement.

Complete step by step answer:
First, we need to simplify the expression, $\sin A+\sin 3A+\sin 5A+\sin 7A$ and $\cos A+\cos 3A+\cos 5A+\cos 7A$. Let’s take the first expression $\sin A+\sin 3A+\sin 5A+\sin 7A$. Rearranging the terms, it can be written as $\sin A+\sin 7A+\sin 3A+\sin 5A$. Using the trigonometric property $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ on the first two and next two terms of the above expression separately, we get
$\Rightarrow 2\sin \left( \dfrac{A+7A}{2} \right)\cos \left( \dfrac{A-7A}{2} \right)+2\sin \left( \dfrac{3A+5A}{2} \right)\cos \left( \dfrac{3A-5A}{2} \right)$
Simplifying the above expression, we get
$\Rightarrow 2\sin 4A\left( \cos (-3A)+\cos (-A) \right)$
Using the property $\cos (-x)=\cos x$ on the above expression, we get
$\Rightarrow 2\sin 4A\left( \cos 3A+\cos A \right)$
Now the second expression, we need to simplify is $\cos A+\cos 3A+\cos 5A+\cos 7A$. Rearranging the terms, it can be written as $\cos A+\cos 7A+\cos 3A+\cos 5A$. Using the property $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ on the first two and next two terms of the above expression, we get
$\Rightarrow 2\cos \left( \dfrac{A+7A}{2} \right)\cos \left( \dfrac{A-7A}{2} \right)+2\cos \left( \dfrac{3A+5A}{2} \right)\cos \left( \dfrac{3A-5A}{2} \right)$
Simplifying the above expression, we get
$\Rightarrow 2\cos 4A\left( \cos (-3A)+\cos (-A) \right)$
Using the property $\cos (-x)=\cos x$ on the above expression, we get
$\Rightarrow 2\cos 4A\left( \cos 3A+\cos A \right)$
We are asked to prove the statement $\dfrac{\sin A+\sin 3A+\sin 5A+\sin 7A}{\cos A+\cos 3A+\cos 5A+\cos 7A}=\tan 4A$. The LHS of the statement is $\dfrac{\sin A+\sin 3A+\sin 5A+\sin 7A}{\cos A+\cos 3A+\cos 5A+\cos 7A}$, and the RHS of the statement is $\tan 4A$.
Let’s simplify the LHS, the numerator of the LHS is $\sin A+\sin 3A+\sin 5A+\sin 7A$, and the denominator of the LHS is $\cos A+\cos 3A+\cos 5A+\cos 7A$. We have already simplified these expressions above, using the simplified forms of these expressions, the LHS can be expressed as
$\Rightarrow \dfrac{2\sin 4A\left( \cos 3A+\cos A \right)}{2\cos 4A\left( \cos 3A+\cos A \right)}$
Canceling out the common factors from the numerator and denominator, we get

& \Rightarrow \dfrac{\sin 4A}{\cos 4A} \\\ & \Rightarrow \tan 4A=RHS \\\ \end{aligned}$$ $$\therefore LHS=RHS$$ Hence, proved. **Note:** To solve these types of questions, one should remember the trigonometric properties. The properties, we used to solve this problem are $$\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$$ and $$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$$. If the LHS has an expression in fraction form then simplifying the numerator and denominator separately is easier to solve.