Question

prove that : (cosx -cosy)square + ( sinx -siny ) square = 4sin square x-y /2.​

Answer 1

Answer:

Step-by-step explanation:

BRAINLIEST BRAINLIEST BRAINLIEST

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

• Here the concept of Trigonometric Identities. In this question mainly we will use 3 identities. Firstly using first identity, we shall simplify the equation to a easier form. Then using the second identity we will find the the main equation to be solved. We will keep on simplifying and then finally using third identity, we will get our answer.

Let's do it !!

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★ Identities Used :-

$\\\;\boxed{\sf{\pink{1.)\;\;\cos\:x\;-\;\cos\:y\;=\;\bf{-2\sin\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin\bigg(\dfrac{x\:-\:y}{2}\bigg)}}}}$

$\\\;\boxed{\sf{\pink{2.)\;\;\sin\:x\;-\;\sin\:y\;=\;\bf{2\cos\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin\bigg(\dfrac{x\:-\:y}{2}\bigg)}}}}$

$\\\;\boxed{\sf{\pink{3.)\;\;\sin^{2}\theta\;+\;\cos^{2}\theta\;=\;\bf{1}}}}$

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★ To Prove :-

$\\\;\bf{\mapsto\;\;\green{(\cos\:x\;-\;\cos\:y)^{2}\;+\;(\sin\:x\;-\;\sin\:y)^{2}\;=\;4\sin^{2}\bigg(\dfrac{x\;-\;2}{2}\bigg)}}$

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★ Solution :-

Given,

$\\\;\bf{\odot\;\;L.H.S.\;=\;(\cos\:x\;-\;\cos\:y)^{2}\;+\;(\sin\:x\;-\;\sin\:y)^{2}}$

And,

$\\\;\bf{\odot\;\;R.H.S.\;=\;4\sin^{2}\bigg(\dfrac{x\;-\;2}{2}\bigg)}$

• Now we shall separately solve LHS to get our answer conveniently.

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{(\cos\:x\;-\;\cos\:y)^{2}\;+\;(\sin\:x\;-\;\sin\:y)^{2}}}$

We know that,

$\\\;\tt{1.)\;\;\cos\:x\;-\;\cos\:y\;=\;-2\sin\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin\bigg(\dfrac{x\:-\:y}{2}\bigg)}$

On applying first identity here, we get

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{\bigg[-2\sin\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg]^{2}\;+\;(\sin\:x\;-\;\sin\:y)^{2}}}$

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{\bigg[(-2)^{2}\sin^{2}\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg]^{2}\;+\;(\sin\:x\;-\;\sin\:y)^{2}}}$

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{\bigg[4\sin^{2}\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg]^{2}\;+\;(\sin\:x\;-\;\sin\:y)^{2}}}$

We know that,

$\\\;\tt{2.)\;\;\sin\:x\;-\;\sin\:y\;=\;2\cos\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin\bigg(\dfrac{x\:-\:y}{2}\bigg)}$

Now applying second identity here, we get

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{\bigg[4\sin^{2}\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg]^{2}\;+\;\bigg[2\cos\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg]^{2}}}$

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{\bigg[4\sin^{2}\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg]^{2}\;+\;\bigg[2^{2}\cos^{2}\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg]}}$

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{\bigg[4\sin^{2}\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg]^{2}\;+\;\bigg[4\cos^{2}\:\bigg(\dfrac{x\:+\:y}{2}\bigg)\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg]}}$

Taking out the common term, we get

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{4\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg[\sin^{2}\bigg(\dfrac{x\;+\;y}{2}\bigg)\;+\;\cos^{2}\bigg(\dfrac{x\;+\;y}{2}\bigg)\bigg]}}$

We know that,

$\\\;\tt{3.)\;\;\sin^{2}\theta\;+\;\cos^{2}\theta\;=\;1}$

Here θ = (x - y)

Now using the third identity here, we get

$\\\;\sf{\rightarrow\;\;L.H.S.\;=\;\bf{4\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)\bigg[1\bigg]}}$

$\\\;\bf{\rightarrow\;\;L.H.S.\;=\;\bf{4\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)}}$

And we know that,

$\\\;\bf{\blue{\rightarrow\;\;L.H.S.\;=\;\bf{4\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)}}}$

$\\\;\bf{\red{\rightarrow\;\;RHS\;=\;\bf{4\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)}}}$

Clearly we get , LHS = RHS.

This means our answer is correct.

$\\\;\bf{\purple{\rightarrow\;\;L.H.S.\;=\;RHS\;=\;\bf{4\sin^{2}\bigg(\dfrac{x\:-\:y}{2}\bigg)}}}$

$\\\;\qquad\qquad\boxed{\tt{Hence,\;\;Proved}}$

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★ More formulas to know :-

• $\\\;\sf{\leadsto\;\;\cos\bigg(\dfrac{\pi}{2}\;+\;x\bigg)\;=\;-\:\sin\:x}$

• $\\\;\sf{\leadsto\;\;\sin\bigg(\dfrac{\pi}{2}\;+\;x\bigg)\;=\;\cos\:x}$

• $\\\;\sf{\leadsto\;\;\cos(\pi\;-\:x)\;=\;-\:\cos\:x}$

• $\\\;\sf{\leadsto\;\;\sin(\pi\;-\:x)\;=\;\sin\:x}$

• $\\\;\sf{\leadsto\;\;\cos(\pi\;+\:x)\;=\;-\:\cos\:x}$

• $\\\;\sf{\leadsto\;\;\sin(\pi\;+\:x)\;=\;-\:\sin\:x}$