Math, asked by Esha109, 10 months ago

Find the value of sin (45° +Ø) - cos (45° - Ø).​

Answers

Answered by PiyushDash
2

Answer:

I hope this will help you.

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AbhijithPrakash: Nice Answer!!
PiyushDash: Thanks
Answered by AbhijithPrakash
7

Answer:

0

Step-by-step explanation:

\sin \left(45^{\circ \:}+\o\right)-\cos \left(45^{\circ \:}-\o\right)

\sin \left(45^{\circ \:}+\o\right)

\mathrm{Use\:the\:following\:identity}:\quad \sin \left(s+t\right)=\cos \left(s\right)\sin \left(t\right)+\cos \left(t\right)\sin \left(s\right)

=\cos \left(45^{\circ \:}\right)\sin \left(\o\right)+\cos \left(\o\right)\sin \left(45^{\circ \:}\right)

\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(45^{\circ \:}\right)=\dfrac{\sqrt{2}}{2}

=\dfrac{\sqrt{2}}{2}\sin \left(\o\right)+\sin \left(45^{\circ \:}\right)\cos \left(\o\right)

\mathrm{Use\:the\:following\:trivial\:identity}:\quad \sin \left(45^{\circ \:}\right)=\dfrac{\sqrt{2}}{2}

=\dfrac{\sqrt{2}}{2}\sin \left(\o\right)+\dfrac{\sqrt{2}}{2}\cos \left(\o\right)

=\dfrac{\sqrt{2}}{2}\sin \left(\o\right)+\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\cos \left(45^{\circ \:}-\o\right)

\cos \left(45^{\circ \:}-\o\right)

\mathrm{Use\:the\:following\:identity}:\quad \cos \left(s-t\right)=\cos \left(s\right)\cos \left(t\right)+\sin \left(s\right)\sin \left(t\right)

=\cos \left(45^{\circ \:}\right)\cos \left(\o\right)+\sin \left(45^{\circ \:}\right)\sin \left(\o\right)

\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(45^{\circ \:}\right)=\dfrac{\sqrt{2}}{2}

=\dfrac{\sqrt{2}}{2}\cos \left(\o\right)+\sin \left(45^{\circ \:}\right)\sin \left(\o\right)

\mathrm{Use\:the\:following\:trivial\:identity}:\quad \sin \left(45^{\circ \:}\right)=\dfrac{\sqrt{2}}{2}

=\dfrac{\sqrt{2}}{2}\cos \left(\o\right)+\dfrac{\sqrt{2}}{2}\sin \left(\o\right)

=\dfrac{\sqrt{2}}{2}\sin \left(\o\right)+\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\left(\dfrac{\sqrt{2}}{2}\cos \left(\o\right)+\dfrac{\sqrt{2}}{2}\sin \left(\o\right)\right)

-\left(\dfrac{\sqrt{2}}{2}\cos \left(\o\right)+\dfrac{\sqrt{2}}{2}\sin \left(\o\right)\right)

\mathrm{Distribute\:parentheses}

=-\left(\dfrac{\sqrt{2}}{2}\cos \left(\o\right)\right)-\left(\dfrac{\sqrt{2}}{2}\sin \left(\o\right)\right)

\mathrm{Apply\:minus-plus\:rules}

+\left(-a\right)=-a

=-\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\dfrac{\sqrt{2}}{2}\sin \left(\o\right)

=\dfrac{\sqrt{2}}{2}\sin \left(\o\right)+\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\dfrac{\sqrt{2}}{2}\sin \left(\o\right)

\mathrm{Simplify}\:\dfrac{\sqrt{2}}{2}\sin \left(\o\right)+\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\dfrac{\sqrt{2}}{2}\sin \left(\o\right)

\dfrac{\sqrt{2}}{2}\sin \left(\o\right)+\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\dfrac{\sqrt{2}}{2}\sin \left(\o\right)

\mathrm{Add\:similar\:elements:}\:\dfrac{\sqrt{2}}{2}\cos \left(\o\right)-\dfrac{\sqrt{2}}{2}\cos \left(\o\right)=0

=\dfrac{\sqrt{2}}{2}\sin \left(\o\right)-\dfrac{\sqrt{2}}{2}\sin \left(\o\right)

\mathrm{Add\:similar\:elements:}\:\dfrac{\sqrt{2}}{2}\sin \left(\o\right)-\dfrac{\sqrt{2}}{2}\sin \left(\o\right)=0

=0


letshelpothers9: nice answer :)
AbhijithPrakash: Thanks!!
PiyushDash: Its too lengthy.
PiyushDash: not need to be lengthy
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