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Last updated date: 01st Dec 2023
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MVSAT Dec 2023

Find the value of ${\text{si}}{{\text{n}}^2}18^\circ - {\text{ co}}{{\text{s}}^2}72^\circ $.

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Hint –To simplify the equation in the given question we use trigonometric identities and covert the sine function into cosine function or vice versa.

Complete step-by-step answer:
${\text{si}}{{\text{n}}^2}18^\circ - {\text{ co}}{{\text{s}}^2}72^\circ $
⟹${\text{si}}{{\text{n}}^2}\left( {90 - 72} \right)^\circ - {\text{ co}}{{\text{s}}^2}72^\circ $ ---- sin (90 – θ) = cosθ, since sine is a periodic function.
⟹${\text{co}}{{\text{s}}^2}72^\circ - {\text{ co}}{{\text{s}}^2}72^\circ $

Hence, ${\text{si}}{{\text{n}}^2}18^\circ - {\text{ co}}{{\text{s}}^2}72^\circ $ = 0

Note: In order to solve this type of question the key is to use the right trigonometric functions and covert one of the trigonometric ratios to another to simplify the equation. Then we solve the obtained function using the trigonometric table.