The equation sin-sin- for all -

The correct option is A True sin (π + x) ⇒sin π cos x + sin x + cos π (cos π = 1 and sin π = 0) ⇒(0) cosx + ( 1) sin x ⇒ sin x Hence, sin(θ + Π) = sin θ for ...

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Standard X

Mathematics

Quadratic Formula

The equation ...

Question

The equation ( $sin θ + Π) = -sin θ$ for all $θ$

True

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False

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Solution

The correct option is A
True

$sin (π + x)$
$\Rightarrow sin π cos x + sin x + cos π$
$(cos π = - 1 and sin π = 0)$
$\Rightarrow (0) cosx + (-1) sin x$
$\Rightarrow -sin x$
$Hence, sin (θ + Π) = -sin θ for all θ$

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Similar questions

The equation sin( $θ + Π$ ) = -sin $θ$ for all acute angle $θ$ .

Q. Find all the values of

θ

satisfying the equation :

sin θ + sin 5 θ = sin 3 θ

such that

0 \leq θ \leq π

The equation, $sin θ = tan θ$ for some $θ$ .

Q. Prove

\sqrt{\frac{1 - sin θ}{1 + sin θ}} + \sqrt{\frac{1 + sin θ}{1 - sin θ}}

= \frac{- 2}{cos θ}, \frac{π}{2} < θ < π

Q. Solve the equation

sin θ + sin 5 θ = sin 3 θ, 0 \leq θ \leq π .

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The equation (sin θ + Π)=-sin θ for all θ

The correct option is A True sin (π + x) ⇒sin π cos x + sin x + cos π (cos π=−1and sin π=0) ⇒(0) cosx + (-1) sin x ⇒-sin x Hence, sin(θ + Π)=-sin θ for all θ

The equation ( $sin θ + Π) = -sin θ$ for all $θ$

The correct option is A
True

$sin (π + x)$
$\Rightarrow sin π cos x + sin x + cos π$
$(cos π = - 1 and sin π = 0)$
$\Rightarrow (0) cosx + (-1) sin x$
$\Rightarrow -sin x$
$Hence, sin (θ + Π) = -sin θ for all θ$