Question

$<i>$ Please help.

$<b>$ I need someone's expert help.

The number 12.

If, cos A + sin A = 1 / 2(cos A - sin A), then find sin^2 A?

Answer 1

(cos A+ sin A) = 1/2*(cos A - sin A)
(cosA+sinA)(cosA-sinA) = 1/2
cos^2 A - sin^2 A = 1/2

Now,
cos^2 A = 1 - sin^2 A
Putting this value in the equation further we get,

1-2sin^2 A = 1/2
2 sin^2 A = 1/2
sin^2 A = 1/4

Hope this helps you !

Answer 2

Answer:

♧♧HERE IS YOUR ANSWER♧♧

Trigonometry is the study of angles and its ratios. This study prescribes the relation amongst the sides of a triangle and angles of the triangle.

There are sine, cosine, tangent, cot, sectant and cosec ratios to an angle of a triangle.

Let me tell you an interesting fact about Trigonometry.

"Triangle" > "Trigonometry"

Remember some formulae now :

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

Want to learn more!

Here it is :

sin(A + B) = sinA cosB + cosA sinB

sin(A - B) = sinA cosB - cosA sinB

cos(A + B) = cosA cosB - sinA sinB

cos(A - B) = cosA cosB + sinA sinB

SOLUTION :

Given :

cosA - sinA = 1

Squaring both sides we get :

(cosA - sinA)² = 1²

=> cos²A - 2cosA sinA + sin²A = 1

=> (cos²A + sin²A) - 2cosA sinA = 1

=> 1 - 2cosA sinA = 1

Now, cancelling 1, we get :

2cosA sinA = 0 .....(i)

Now,

(cosA + sinA)²

= cos²A + 2cosA sinA + sin²A

= cos²A + sin²A + 2×0, by (i)

= 1

Since, sin²A + cos²A always values 1,

cosA + sinA = 1, ≠ -1.

Therefore?

cosA + sinA = 1

Hence, proved.

♧♧HOPE THIS HELPS YOU♧♧