calculate the sum sin x+square root(1-sin^2x)
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We have to calculate sin x + sqrt ( 1 - (sin x)^2)
sin x + sqrt ( 1 - (sin x)^2)
=> sin x + sqrt ((cos x)^2)
=> sin x + cos x
This sum can be changed to many other forms but the basic result is sin x + cos x
Therefore the sum = sin x + cos x
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We have the fundamental Identity: cos^2(x) + sin^2(x) = 1
We can manipulate it in the given way -
cos^2(x) = 1 - sin^2(x)
=> cos(x) = ± √( 1 - sin^2(x))
so, we can write sin x + sqrt(1-sin^2x) as
= sin(x) ± cos(x)
The sign of ± cos(x) depends upon the value of x.
To calculate the expression we'll have to transform the sum into a product. The terms of the su are not like trigonometric functions.
We'll re-write the second term: sqrt(1-sin^2x) = sqrt (cos x)^2
sin x + sqrt(1-sin^2x) = sin x + sqrt (cos x)^2
sin x + sqrt(1-sin^2x) = sin x + cos x
We'll express the function cosine, depending on the function sine.
cos x= sin (90-x)
The expression will become:
sin x + cos x = sinx + sin (90-x)
Now we can transform the expression into a product:
sin x + cos x = 2 sin (x+90-x)/2*cos (x-90+x)/2
sin x + cos x = 2 sin 45*cos [-(90-2x)/2]
sin x + cos x = 2* (sqrt2/2)*cos (45-x)
sin x + cos x = sqrt 2*(cos 45*cos x + sin 45*sin x)
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