sin a= o/h, in the 40 5,40 5 suitable angled triangle the facets = x and the hypotenuse = x*root(2) subsequently sin 40 5 might equivalent to x/x*root(2) = a million/root(2) that's a relentless fee for sin and cos 40 5, in the 30, 60 triangle the area opposite to 30 is x, to 60 is x*root(3), the hypotenuse is 2x so sin 30 = x/2x = a million/2 = cos 60 (sin a = cos ninety-a) and cos 30 = (x*root(3))/2x = root(3)/2 = sin60 then use the sum rule: Sum and distinction formula sin(A+B)=sin A cos B + cos A sin B sin(A-B)=sin A cos B - cos A sin B cos(A+B)=cos A cos B - sin A sin B cos(A-B)=cos A cos B + sin A sin B so sin70 = sin(40 5+30) = sin45cos30+cos45sin30 = (a million/root2 * root3/2) + (a million/root2 * a million/2) = root3/2root2 + a million/2root2 = (a million+root3) / 2root2 use a calculator to verify the end result
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cos 75° cos 45°- sin 75°sin 45
= cos (75+45)
= cos 120
= - cos 60
= - 1/2 >>... A
sin a= o/h, in the 40 5,40 5 suitable angled triangle the facets = x and the hypotenuse = x*root(2) subsequently sin 40 5 might equivalent to x/x*root(2) = a million/root(2) that's a relentless fee for sin and cos 40 5, in the 30, 60 triangle the area opposite to 30 is x, to 60 is x*root(3), the hypotenuse is 2x so sin 30 = x/2x = a million/2 = cos 60 (sin a = cos ninety-a) and cos 30 = (x*root(3))/2x = root(3)/2 = sin60 then use the sum rule: Sum and distinction formula sin(A+B)=sin A cos B + cos A sin B sin(A-B)=sin A cos B - cos A sin B cos(A+B)=cos A cos B - sin A sin B cos(A-B)=cos A cos B + sin A sin B so sin70 = sin(40 5+30) = sin45cos30+cos45sin30 = (a million/root2 * root3/2) + (a million/root2 * a million/2) = root3/2root2 + a million/2root2 = (a million+root3) / 2root2 use a calculator to verify the end result
sin(α + ß) = sin(α)cos(ß) + sin(ß)cos(α)
cos(α + ß) = cos(α)cos(ß) - sin(α)sin(ß)
cos(75° + 45°) = cos(75°)cos(45°) - sin(75°)sin(45°)
cos(75° + 45°) = cos(120°) = -cos(60°) = -½