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it equals 4

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Q: If θ is an acute angle and sin θ equals 2 then cos θ?
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If θ is an acute angle and sin θ equals 12 then cos θ equals?

The value of sine and cosine varies between -1 and 1, it never can be 12. If you mean .12, then we have: sin x = .12 x = sin-1 .12 x = 6.89 degrees So that, cos x = cos 6.89 = .99


X is acute and sin x equals cos x Find cos?

x = 45 degrees sin(x) = cos(x) = 1/2 sqrt(2)


How do you calculate trigonometric ratios of obtuse angles?

If ø is an obtuse angle then (180 - ø) is an acute angle and: sin ø = sin (180 - ø) cos ø = -cos (180 - ø) tan ø = -tan (180 - ø)


Write down the acute angle whose sine is equal to Cos 70?

cos70=sin20 so angle is 20, because cosA=sin(90-A)


What is this expression as the cosine of an angle cos30cos55 plus sin30sin55?

cos(30)cos(55)+sin(30)sin(55)=cos(30-55) = cos(-25)=cos(25) Note: cos(a)=cos(-a) for any angle 'a'. cos(a)cos(b)+sin(a)sin(b)=cos(a-b) for any 'a' and 'b'.


When does cos x equal -sin x?

The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).


Given that Sin x equals 0.8 and x is acute write down the value of Cos x?

cos2x = 1 - sin2x = 1 - 0.64 = 0.36 So cos x = +/- 0.6 Since x is acute, cos x is +ve, so cos x = 0.6


How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?

sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)


How do you verify the identity of cos θ tan θ equals sin θ?

To show that (cos tan = sin) ??? Remember that tan = (sin/cos) When you substitute it for tan, cos tan = cos (sin/cos) = sin QED


How would you prove left cosA plus sinA right times left cos2A plus sin2A right equals cosA plus sin3A?

You need to make use of the formulae for sin(A+B) and cos(A+B), and that cos is an even function: sin(A+B) = cos A sin B + sin A cos B cos(A+B) = cos A cos B - sin A sin B cos even fn → cos(-x) = cos(x) To prove: (cos A + sin A)(cos 2A + sin 2A) = cos A + sin 3A The steps are to work with the left hand side, expand the brackets, collect [useful] terms together, apply A+B formula above (backwards) and apply even nature of cos function: (cos A + sin A)(cos 2A + sin 2A) = cos A cos 2A + cos A sin 2A + sin A cos 2A + sin A sin 2A = (cos A cos 2A + sin A sin 2A) + (cos A sin 2A + sin A cos 2A) = cos(A - 2A) + sin(A + 2A) = cos(-A) + sin 3A = cos A + sin 3A which is the right hand side as required.


Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?

[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,


Determine the measure of angle A if 0 degrees is smaller than or equal to A which is smaller than or equal to 90 degrees and cos125 degrees equals negative cos A?

cos(125) = cos(180 - 55) = cos(180)*cos(55) + sin(180)*sin(55) = -cos(55) since cos(180) = -1, and sin(180) = 0 So A = 55 degrees.