No, (sinx)^2 + (cosx)^2=1 is though
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
1. Anything divided by itself always equals 1.
Since the word 'equals' appears in your questions it might be what is called a trigonometric identity, in other words a statement about a relationship between various trigonometric values.
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The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
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1. Anything divided by itself always equals 1.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
2 x cosine squared x -1 which also equals cos (2x)
Since the word 'equals' appears in your questions it might be what is called a trigonometric identity, in other words a statement about a relationship between various trigonometric values.
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
According to the Pythagorean identity, it is equivalent to sin2theta.
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