Question

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

Answer 1

Hint: 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
\[\begin{align}
  & \Rightarrow \dfrac{\sin 4A}{\cos 4A} \\
 & \Rightarrow \tan 4A=RHS \\
\end{align}\]
\[\therefore LHS=RHS\]
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.