Write down the values of left side trigonometric functions -sin -160-2 -960-x -160- -160-sin -160-2 -960-x -160- -160-cos -160-2 -960-x -160- -160-cos -160-2 -960-x -160- -160-sin -160- -960-x -160- -160-sin -160-xsin -160- -960-x -160- -160-sin -160-xcos -160- -960-x -160- -160-cos -160-xcos -160- -960-x -160- -160-cos -160-x

Dear student sin2 #960; x= sinxsin2 #960;+x=sinxcos2 #960;+x=cosxcos2 #960; x=cosxsin #960; x=sinxsin #960;+x= sinxcos #960;+x= cosxcos #960; x= ...

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Byju's Answer

Standard XII

Mathematics

Property 7

Write down th...

Question

Write down the values of left side trigonometric functions .

$\sin (2 π - x) = ? \sin (2 π + x) = ? c o s (2 π + x) = ? \cos (2 π - x) = ? \sin (π - x) = \sin x \sin (π + x) = - \sin x \cos (π + x) = - \cos x \cos (π - x) = - \cos x$

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Solution

Dear student
$\sin (2 π - x) = - sinx \sin (2 π + x) = sinx \cos (2 π + x) = cosx \cos (2 π - x) = cosx \sin (π - x) = sinx \sin (π + x) = - sinx \cos (π + x) = - cosx \cos (π - x) = - cosx$
Regards

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Similar questions

The number of distinct real roots of $∣ ∣ ∣ ∣ \begin{matrix} s i n x & c o s x & c o s x c o s x & s i n x & c o s x c o s x & c o s x & s i n x \end{matrix} ∣ ∣ ∣ ∣ = 0$ in the interval $- \frac{π}{4} \leq x \leq \frac{π}{4}$ is

(a) 0
(b) 2
(c) 1
(d) 3

Q. The number of distinct real roots of

∣ ∣ ∣ ∣ \begin{matrix} sin x & cos x & cos x cos x & sin x & cos x cos x & cos x & sin x \end{matrix} ∣ ∣ ∣ ∣ = 0

in the interval

- \frac{π}{4} \leq x \leq \frac{π}{4}

$if s i n^{- 1} x + s i n^{- 1} y = \frac{2 π}{3} and c o s^{- 1} x - c o s^{- 1} y = \frac{π}{6}, then x =$

Q. Solve for

x

and

y

;

{sin}^{- 1} x + {sin}^{- 1} y = \frac{2 π}{3}

and

{cos}^{- 1} x - {cos}^{- 1} y = \frac{π}{3}

$\frac{d}{d x} (\sqrt{c o s \sqrt{x}}) =$

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Write down the values of left side trigonometric functions . sin 2π-x = ?sin 2π+x = ?cos 2π+x = ?cos 2π-x = ?sin π-x = sin xsin π+x = -sin xcos π+x = -cos xcos π-x = -cos x

Dear student sin2π-x=-sinxsin2π+x=sinxcos2π+x=cosxcos2π-x=cosxsinπ-x=sinxsinπ+x=-sinxcosπ+x=-cosxcosπ-x=-cosx Regards

Write down the values of left side trigonometric functions .

$\sin (2 π - x) = ? \sin (2 π + x) = ? c o s (2 π + x) = ? \cos (2 π - x) = ? \sin (π - x) = \sin x \sin (π + x) = - \sin x \cos (π + x) = - \cos x \cos (π - x) = - \cos x$

Dear student
$\sin (2 π - x) = - sinx \sin (2 π + x) = sinx \cos (2 π + x) = cosx \cos (2 π - x) = cosx \sin (π - x) = sinx \sin (π + x) = - sinx \cos (π + x) = - cosx \cos (π - x) = - cosx$
Regards