Sin 38-cos 52-a 0b 1c 32d 23

Sin38 #176; cos52 #176;=sin90 #176; 52 #176; cos52 #176;=cos52 #176; cos52 #176; #160; #160; #160; #160; #160; #160; #160; #160; ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

sin 38∘-cos 5...

Question

sin 38° – cos 52° = ?
(a) 0

(b) 1

(c) $\frac{\sqrt{3}}{2}$

(d) $\frac{2}{\sqrt{3}}$

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Solution

$\sin 38 ° - \cos 52 ° = \sin (90 ° - 52 °) - \cos 52 ° = \cos 52 ° - \cos 52 ° (∵ \sin (90 ° - θ) = \cos θ) = 0 Hence, the correct option is (a) .$

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Similar questions

Fill in the blanks
i) $t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ}$ = ___
ii) $c o s 38^{\circ} . c o s 52^{\circ} - s i n 38^{\circ} . s i n 52^{\circ}$ = ___

Q. Simplify

cos 38^{\circ} cos 52^{\circ} - sin 38^{\circ} sin 52^{\circ}

Q. The value of

t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ}

and

c o s 38^{\circ} . c o s 52^{\circ} - s i n 38^{\circ} . s i n 52^{\circ}

is equal to

Q. show that

cos 38^{\circ} cos 52^{\circ} - sin 38^{\circ} sin 52^{\circ} = 0

Q. Show that:

c o s 38^{\circ} c o s 52^{\circ} - s i n 38^{\circ} s i n 52^{\circ} = 0

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sin 38° – cos 52° = ? (a) 0 (b) 1 (c) 32 (d) 23

sin38°-cos52°=sin90°-52°-cos52°=cos52°-cos52° ∵ sin90°-θ=cosθ=0Hence, the correct option is a.

sin 38° – cos 52° = ?
(a) 0

(b) 1

(c) $\frac{\sqrt{3}}{2}$

(d) $\frac{2}{\sqrt{3}}$

$\sin 38 ° - \cos 52 ° = \sin (90 ° - 52 °) - \cos 52 ° = \cos 52 ° - \cos 52 ° (∵ \sin (90 ° - θ) = \cos θ) = 0 Hence, the correct option is (a) .$