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Q: What is the cos of angle B 5 13 12?
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What is the cos of angle 'a' with the measurements 5 13 12?

Use Cosine Rule a^(2) = b^(2) + c^(2) - 2bcCosA Algebrically rearrange CosA = [a^(2) - b^(2) - c^(2)] / -2bc Substitute CosA = [13^(2) - 12^(2) - 5^(2)# / -2(12)(5) CosA = [ 169 - 144 - 25] / -120 Cos)A) = [0] / -120 CosA = 0 A = 90 degrees (the right angle opposite the hypotenuse)/ However, If 'A' is the angle between '12' & '13' then 'a' is the side '5' Hence (Notice the rearrangement of the numerical values). CosA = [5^(2) - 12^(2) - 13^(2) ] / -2(12)(13) CosA = [ 25 - 144 -169] / -312 CosA = [ -288[/-312 CosA = 288/312 A = Cos^(-1) [288/312] A = 22.61986495.... degrees.


What is the sin of angle B if the angles are 5 12 and 13?

This is a classic Pythagorean triangle. Although you have given the side lengths, you have NOT given a letter to correspond , with the given side. However, Let 12 be the adjacentr side (base) Let '5' be the opposite side ( perpendicular ) Let '13' by the hypotenuse. Sin(Angle) = opposite / hypotenuse = 5/13 Angle = Sin^(-1) 5/13 = 22.619... degrees. NB This is the angle between the hypotenuse and the base(adjacent) Now 'swopping' things around , we take the angle between the hypotenuse and the perpendicular (opposite) . This now becomes perpendicular(adjacent) and the base becomes the opposite. Hence Sin(angle) = 12/13 Angle = Sin^(-1) 12/13 = 67.380.... degrees. The angle at the 'top' of the triangle. Verification. ' 90 + 67.380... + 22.619... = 180 ( allow for calculator decimals).


Express cos4x sin3x in a series of sines of multiples of x?

The best way to answer this question is with the angle addition formulas. Sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and cos(a + b) = cos(a)cos(b) - sin(a)sin(b). If you compute this repeatedly until you get sin(3x)cos(4x) = 3sin(x) - 28sin^3(x) + 56sin^5(x) - 32sin^7(x).


How do you confirm this trig identity?

sin^5 2x = 1/8 sin2x (cos(8x) - 4 cos(4x)+3)


What is the reference angle for an angle with a measure of 175?

5 dgs