Question

how sin(90+theta) and sin(90-theta) can be cos(theta)​

Answer 1

Step-by-step explanation:

In a triangle,

let the Angles be x, y and z

where one of the Angles are 90° and sum of all angles will be 180°

(Remember that trigonometric functions are used for right triangles!!!)

So a 90° is a must.

Now,

Let y = 90°

If we draw a triangle ABC

where ∠A = x, ∠B = y = 90°, ∠C = z

Then,

Sin x = BC/AC

Sin z = AB/AC

Now, lastly we need to know one more thing,

x + y + z = 180° (Angle Sum Property)

x + 90° + z = 180°

x = 180° - 90° - z

x = 90° - z

Similarly, we can prove,

z = 90° - x

Now, let's prove our Question,

Sin x = BC/AC

Also,

Sin x = Sin (90° - z)

Then, comparing both the equations,

Sin (90° - z) = BC/AC ----- 1

And,

Cos z = BC/AC ------ 2

From eq.1 and eq.2 we get

Sin (90° - z) = Cos z = BC/AC

Hence,

Sin (90° - z) = Cos z

Similarly,

We can prove Cos x = Sin (90° - x)

Cos x = AB/AC

And,

Sin z = Sin (90° - x)

Also,

Sin z = AB/AC

Then,

Sin (90° - x) = Cos x = AB/AC

So,

Cos x = Sin (90° - x)

Similarly,

Cos x = Sin (90° + x) and it needs a higher level of education about trigonometry but it is valid, so you can use it.

Hope it answers you questions...All the best