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Question: Solve the expression \(\sin 10^\circ + \sin 20^\circ + \sin 40^\circ + \sin 50^\circ = \sin 70^\circ...

Solve the expression sin⁡10∘+sin⁡20∘+sin⁡40∘+sin⁡50∘=sin⁡70∘+sin⁡80∘\sin 10^\circ + \sin 20^\circ + \sin 40^\circ + \sin 50^\circ = \sin 70^\circ + \sin 80^\circ .

Explanation

Solution

In the given question, we are given a problem in the form of addition of sine functions. We can solve the given problem by using formula of addition of two sin⁡\sin functions, sin⁡A+sin⁡B\sin A + \sin B. Take sin⁡50∘+sin⁡10∘\sin 50^\circ + \sin 10^\circ and sin⁡40∘+sin⁡20∘\sin 40^\circ + \sin 20^\circ , simplify these two using the formula. Substitute values of trigonometric ratios wherever required and at the end convert the cosine function of trigonometry into the sine function of trigonometry.
Formula used: sin⁡A+sin⁡B\sin A + \sin B

Complete step-by-step solution:
We have, sin⁡10∘+sin⁡20∘+sin⁡40∘+sin⁡50∘=sin⁡70∘+sin⁡80∘\sin 10^\circ + \sin 20^\circ + \sin 40^\circ + \sin 50^\circ = \sin 70^\circ + \sin 80^\circ
Here, LHS: sin⁡10+sin⁡20+sin⁡40+sin⁡50\sin 10 + \sin 20 + \sin 40 + \sin 50
We know the formula of addition of two sin⁡\sin functions, sin⁡A+sin⁡B\sin A + \sin B.
sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
Let us take the given problem in the form of addition of two sin⁡\sin functions.
(sin⁡50+sin⁡10)+(sin⁡40+sin⁡20)\left( {\sin 50 + \sin 10} \right) + \left( {\sin 40 + \sin 20} \right)
Apply the above written formula,
⇒(2sin⁡(50+102)cos⁡(50−102))+(2sin⁡(40+202)cos⁡(40−202))\Rightarrow \left( {2\sin \left( {\dfrac{{50 + 10}}{2}} \right)\cos \left( {\dfrac{{50 - 10}}{2}} \right)} \right) + \left( {2\sin \left( {\dfrac{{40 + 20}}{2}} \right)\cos \left( {\dfrac{{40 - 20}}{2}} \right)} \right)
On Addition and subtraction of angles, we get
⇒(2sin⁡(602)cos⁡(402))+(2sin⁡(602)cos⁡(202))\Rightarrow \left( {2\sin \left( {\dfrac{{60}}{2}} \right)\cos \left( {\dfrac{{40}}{2}} \right)} \right) + \left( {2\sin \left( {\dfrac{{60}}{2}} \right)\cos \left( {\dfrac{{20}}{2}} \right)} \right)
On division of angles inside the brackets, we get
⇒2sin⁡30cos⁡20+2sin⁡30cos⁡10\Rightarrow 2\sin 30\cos 20 + 2\sin 30\cos 10
Take 2sin⁡30∘2\sin 30^\circ as common
⇒2sin⁡30(cos⁡20+cos⁡10)\Rightarrow 2\sin 30\left( {\cos 20 + \cos 10} \right)
We know the value of, sin⁡30∘=12\sin 30^\circ = \dfrac{1}{2}
On substituting value of sin⁡x\sin x, we get
⇒2×12(cos⁡20+cos⁡10)\Rightarrow 2 \times \dfrac{1}{2}\left( {\cos 20 + \cos 10} \right)
⇒cos⁡20+cos⁡10\Rightarrow \cos 20 + \cos 10
By using the trigonometric cosine function cos⁡(90−θ)=sin⁡θ\cos \left( {90 - \theta } \right) = \sin \theta we get
⇒cos⁡(90−70)+cos⁡(90−80)\Rightarrow \cos \left( {90 - 70} \right) + \cos \left( {90 - 80} \right)
⇒sin⁡70+sin⁡80\Rightarrow \sin 70 + \sin 80
The value we obtained after solving the LHS is equal to RHS value, hence proved LHS = RHS.

Note: To solve these type of questions we should know all the required values of standard angles say, 0∘0^\circ , 30∘30^\circ , 60∘60^\circ , 90∘90^\circ , 180∘180^\circ , 270∘270^\circ , 360∘360^\circ respectively for each trigonometric term such as sin⁡\sin , cos⁡\cos , tan⁡\tan , cot⁡\cot , sec⁡\sec , cos⁡ec\cos ec, etc. We should take care of the calculations so as to be sure of our final answer. In the given question we added two sin⁡\sin functions. Similarly we can subtract two sin⁡\sin functions using formula of subtraction of two sin⁡\sin functions. Also, subtract or add two cos⁡\cos functions by using formulae. Such as,
sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
We use them according to the given problem.