Question: Sin x + Cos x = 1 ; Solve for x.
Sin x + Cos x = 1 //Original Question
[Sin x + Cos x]2 = [1]2 //Square both sides
Sin2 x + 2 Sin x Cos x + Cos2 x = 1 // Expand.
1 + 2 Sin x Cos x = 1 //Note that sin2 x + cos2 x = 1 from the trigonometry identities. Look for proof below.
2 Sin x Cos x = 0 // Subtract 1 from both sides
So either
1) Sin x = 0
Looking at the unit circle, then this means x = 180ko where k E I
2) Cos x = 0
Looking at the unit circle, then this means x = 90 + 180k where k E I
3) 2 = 0
No solutions for this case.
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* So that's all x for which Sin x + Cos x = 1. *
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Proof for Identity used to solve this question:
Sin2 x + Cos2 x = 1
Imagine a right-angled triangle with an angle x and sides (A adjacent, O opposite, H hypotenuse):
_____.
____...
_H_.....
__....... O
_.........
..x.......
A
Then you know from the Pythagorean Theorem that A2 + O2 = H2
You also know from the definition of Cos and Sin that:
Sin x = O / H
Cos x = A / H
Now take the sum of the squares of both.
Sin2 x + Cos2 x
= [O / H]2 + [A / H]2
= O2 / H2 + A2 / H2
= ( O2 + A2 ) / H2
= ( H2 ) / H2
= 1
Q.E.D.
That's it.
Good Luck!
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cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x