tanx = 0.5 find sinx and cosx

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tanx = 0.5

Therefore sin x = tanx*cosx = tanx/ secx = tanx/sqrt(1+tan^2x).

So sinx = 0.5/sqrt(1+0.5^2) = 0.5/+(1.25)^(1/2) = 0.447213595 in 1st quadrant.

sinx = 0.5/-sqrt(1.0.5^2) = -0.447213595 in 3rd quadrant.

Similarly cosx = 1/secx = 1/sqrt(1+tan^2x) or 1/-srt(1+tan^2x)

cosx = 1/sqrt(1+0.5^2) = 0.894427191 or -0.894427191

We know from the fundamental formula of trigonometry that:

(sin x)^2 + (cos x)^2 = 1

If we divide by (cos x)^2, both sides, we'll get:

(sin x/cos x)^2 + 1 = 1/(cos x)^2

But sin x / cos x = tan x

(tan x)^2 + 1 = 1/(cos x)^2

But tan x = 0.5 = 50/100 = 1/2

(1/2)^2 + 1 = 1/(cos x)^2

5/4 = 1/(cos x)^2

We'll cross multiply and we'll get:

4 = 5(cos x)^2

We'll divide by 5:

4/5 = (cos x)^2

cos x = +/- 2/sqrt5

**cos x = +/- 2sqrt5/5**

sin x = +/-sqrt(1 - 4/5)

**sin x = +/-sqrt5/5**

**Conclusion: The values of sine and cosine have to be both positive, or both negative, in order to obtain the positive value of tan x = 0.5**

tanx = 1/2

But we know that :

tanx = sinx/cosx = 1/2

==> 2sinx = cosx..........(1)

Also , we know that:

sin^2 x + cos^2 x = 1

==> cos x = sqrt(1-sin^2 x)..........(2)

Now substitute in (1)

==> 2sin x = sqrt(1-sin^2 x)

Square both sides:

==> 4sin^2 x = 1-sin^2 x

==> 5sin^2 x = 1

Divide by 5

==> sin^2 x = 1/5

==> sin x= 1/sqrt5= sqrt5/5

==> **sin x = sqrt5/5**

Now substitute in (1):

cosx = 2sinx

= 2(sqrt5/5)

= 2sqrt4/5

==> **cos x = 2sqrt5/5**

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