Question

please solve the question correctly correctly.......DON'T SPAM......​

Answer 1

Answer:

your answer attached in the photo

Answer 2

Question:

if 3 cosα -4sinα = 5

so prove that 3 sinα + 4 cos α = 0

Trignometric Formulas :

• sinA cos B + sinB cosA = sin( A+ B)
• sinAcosB - sinBcosA =sin(A-B)
• cosAcosB -sinAsinB=cos( A+B)
• cosAcosB+sinAsinB =cos(A-B)
• sin²A + cos²A = 1
• sec²A - tan²A = 1
• cosec²A - cot²A = 1
• sin2A = 2 sinA cosA
• cos2A = cos²A - sin²A
• tan2A = 2 tanA / (1 - tan²A)

Solution :

Given : 3 cosα -4sinα = 5

Now , divide both sides by 5

$\implies \: \frac{3}{5} \cos( \alpha ) - \frac{4}{5} \sin( \alpha ) = 1$

let sin x = $\frac{3}{5}$

⇒cosx = $\frac{4}{5}$

$sinx \: cos \alpha - cosx \: sin \alpha = 1$

⇒sin(x-α) = 1

we know that sin²A + cos²A = 1

$\implies \: \cos(x - \alpha ) = \sqrt{1 - \sin {}^{2} (x - \alpha ) }$

⇒cos(x-α) = 0

________________________

we have to prove :3 sinα + 4 cos α = 0

LHS :

3 sinα + 4 cos α

$= 5( \frac{3}{5} \sin \alpha - \frac{4}{5} \cos \alpha )$

$= 5(sinx \: sin \alpha - cosx \:cos \alpha )$

$= - 5(cosx \: cos \alpha - sinx \: sin \alpha )$

$= - 5 \times cos(x - \alpha )$

put the value of cos (x-α)

$= 0$

RHS = 0

⇒LHS = RHS

$\huge{\bold{ Hence \: Proved }}$