Q: How do you write the expression sin 37 in terms of cosine?

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Thanks to the pre-existing addition and subtraction theorums, we can establish the identity:sin(a+b) = sin(a)cos(b)+sin(a)cos(b)Then, solving this, we getsin(a+b) = 2(sin(a)cos(b))sin(a)cos(b) = sin(a+b)/2a=b, sosin(a)cos(a) = sin(a+a)/2sin(a)cos(a) = sin(2a)/2Therefore, the answer is sin(2a)/2.

Every angle has a sine and a cosine. The sine of 35 degrees is 0.57358 (rounded) The cosine of 35 degrees is 0.81915 (rounded)

Because it is not always convenient to convert to convert sin to cos when working with adjacent sides.

it is the same as a sin function only shifted to the left pi/2 units

We write sin x * sin x = sin2 x

Related questions

cos(x) = sin(pi/2-x) = -sin(x-pi/2)

csc(x)*{sin(x) + cos(x)} = csc(x)*sin(x) + csc(x)*cos(x) =1/sin*(x)*sin(x) + 1/sin(x)*cos(x) = 1 + cot(x)

sin = opp/hyp cos = adj/hyp tan = opp/adj

the only close answer i know is: eix = cos(x)+i*sin(x) where i is imaginary unit

cos = sqrt(1 - sin^2)

No, it does not.

cos(35)sin(55)+sin(35)cos(55) If we rewrite this switching the first and second terms we get: sin(35)cos(55)+cos(35)sin(55) which is a more common form of the sin sum and difference formulas. Thus this is equal to sin(90) and sin(90)=1

f(x) = Cos(x) f'(x) = -Sin(x) Conversely f(x) = Sin(x) f'(x) = Cos(x) NB Note the change of signs.

sin 0 = 0 cos 0 = 1

csc(x) = 1/sin(x) = +/- 1/sqrt(1-cos^2(x))

Generally, the derivative of sine is cosine.

Sin, cosine, and tangent are considered the three main of trigonometry, commonly written as sin, cos, and tan. sin(θ) = O/H cos(θ) = A/H tan(θ) = O/A Where O is opposite Where H is Hypotenuse Where A is Adjacent To assist further in understanding: http://www.mathsisfun.com/sine-cosine-tangent.html