Question

) Prove that geometrically cos(x + y) = cos x cos y - sin x sin y and hence find the
value of cos[π÷2 + x]​

Answer 1

Answer:

${ \huge \orange{ \underline{answer = - \sin(x) }}}$

$\mathfrak{ \large \red{ \underline{question}}} \\ \implies { \large{find \: the \: value \: of \: \cos( \frac{\pi}{2 } + x) }} \\ \mathfrak{ \large \green{ \underline{formula}}} \\ \implies{ \large{ \cos(x + y) = \cos(x) \cos(y) - \sin(x) \sin(y) }} \\ \\ \mathfrak{ \large \blue{ \underline{solution}}} \\ \implies{ \large{ \cos( \frac{\pi}{2} ) \cos(x) - \sin( \frac{\pi}{2} ) \sin(x) }} \\ \implies{ \large{o \times \cos(x) - 1 \times \sin(x) }} \\ \implies{ \large{0 - \sin(x) }} \\ \implies{ \large{ - \sin(x) }} \\ { \large \orange{ \underline{answer = - \sin(x) }}}$

Answer 2

$\huge\mathbb{\underline{\underline{Question}}}$

$\tt{Find\:the\:the\:value\:of}$

$\cos( \frac{\pi}{2} + x)$

$\huge\mathbb{\underline{\underline{Formula\:used}}}$

$\cos(x + y) = \cos(x) \cos(y) - \sin(x ) \sin(y)$

$\huge\mathbb{\underline{\underline{Solution}}}$

$\cos( \frac{\pi}{2} ) \cos(x ) - \sin( \frac{\pi}{2} ) \sin(x) \\ →0 \times \cos(x) - 1 \times \sin(x) \\ →0 - \sin(x) \\ → - \sin(x)$

$\tt{Answer\:is}$$\fbox{-sin(x)}$