Question

If,
$a \cos\alpha + b \sin\alpha = c$
Then prove that
$a \sin \alpha - b \cos \alpha = { + }{ - } \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} }$
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Answer 1

Hey !!!
acos@ + bsin@ = c ------1)

and , asin@ - bcos@ = +-√a²+ b² - c² proofe

First of all adding (acos@ + bsin@ ) and (asin@ - bcos@) and squaring both .

the we get ,

(acos@-bsin@ )² + (asin@+bcos@)²

= a²cos²@+b²sin²@-2acos@×b²sin@ +a²sin²@+b²cos²@ -2asin@×bcos@

= a²cos²@ + a²sin²@ +b²sin²@ + b²cos²@
↪ Rearranging the term

= a²(cos²@ + sin²@) + b²(sin²@ + cos²@)

= a² + b² -------2)

now , again adding ( acos@-bsin@) asin@+bcos@) and squaring both

(acos@-bsin@)² +(asin@+bcos@)² = a²+b²
↪from equation 2)

c² + (asin@ + bcos@)² = a²+b²
↪from equation 1)

(asin@ + bcos@)² = a²+b² -c²

asin@+bcos@ = +-√a²+b² -c² prooved

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Hope it helps you !!!

@Rajukumar111@@@@

Answer 2

a cosα + b sinα = c

Squaring on both sides.

(a cosα + b sinα)² = c²

a² cos²α +b² sin²α + 2(a cosα × b sinα)= c² 

From this using formula sin²α + cos²α = 1 

Then, a²(1-sin²α) + b² (1-cos²α) = c² - 2(a cosα × b sinα)

a² - a² sin²α + b² - b² cos²α = c² - 2(a cosα × b sinα)

a² + b² - c² = a² sin²α + b² cos²α - 2(a cosα × b sinα)

Rearranging (a cosα × b sinα) = (a sinα × b cosα)

a² + b² - c² = (a sinα - b cosα)²

a sinα - b cosα = +√ a² + b² - c² ,  -√a² + b² - c² 

Hence proved.
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