Prove that -3cos23 - - sin23 -2 - cos53 -

Taking L.H.S. #160; √(3)cos(23^∘) sin(23^∘2 = √(3)2cos(23^∘) 12sin(23^∘) = cos (30^∘)cos(23^∘) sin (30^∘)sin(23^∘) =cos(30^∘+23^∘) = cos (53^∘) ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

Prove that ...

Question

Prove that $\frac{\sqrt{3} cos 23^{\circ} - sin 23^{\circ}}{2} = cos 53^{\circ}$

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Solution

Taking L.H.S.
$\frac{\sqrt{3} c o s (23^{\circ}) - s i n (23^{\circ}}{2} = \frac{\sqrt{3}}{2} c o s (23^{\circ}) - \frac{1}{2} s i n (23^{\circ})$
$= c o s (30^{\circ}) c o s (23^{\circ}) - s i n (30^{\circ}) s i n (23^{\circ})$
$= c o s (30^{\circ} + 23^{\circ}) = c o s (53^{\circ}) = R . H . S .$

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$\frac{1}{4}$ $[\sqrt{3} c o s 23^{\circ} - s i n 23^{\circ}]$ =

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is equal to

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\frac{cos 67^{\circ}}{sin 23^{\circ}}

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Prove that √3cos23∘−sin23∘2=cos53∘

Taking L.H.S. √3cos(23∘)−sin(23∘2=√32cos(23∘)−12sin(23∘)=cos(30∘)cos(23∘)−sin(30∘)sin(23∘)=cos(30∘+23∘)=cos(53∘)=R.H.S.

Prove that $\frac{\sqrt{3} cos 23^{\circ} - sin 23^{\circ}}{2} = cos 53^{\circ}$

Taking L.H.S.
$\frac{\sqrt{3} c o s (23^{\circ}) - s i n (23^{\circ}}{2} = \frac{\sqrt{3}}{2} c o s (23^{\circ}) - \frac{1}{2} s i n (23^{\circ})$
$= c o s (30^{\circ}) c o s (23^{\circ}) - s i n (30^{\circ}) s i n (23^{\circ})$
$= c o s (30^{\circ} + 23^{\circ}) = c o s (53^{\circ}) = R . H . S .$