Why Am I Struggling with Sin and Cos Problems Before My Math Exam?

[Warwick]

I'm doing a review for my math class and came upon some that I cannot figure out.I don't know why I cannot get the right answer. I know all the rules, I've had tests on this stuff and received A's but these are different for some reason and I don't know why. If anyone can do even one of them I would so greatly appreciate it! my exam is tomarrow thanks




let sin(x)=5/13, cos(x)=12/13, sin(y)=4/5 and y is in Quadrant II and 0<or=to x <2pi

81. sin(x-y)=____

83. sin 2x =___

84. sin x/2 =___

now the rules above don't apply to these below

85. sin(pheta)=2/3 and 0<pheta<90 then cos2(pheta)=___

86.sin(2cos^-1(1/4))=___

 

[arildno]

1.
Determine the cosine value of y:
You are given that y is in Quadrant 2, which means that the cosine value of y is negative (Agreed?)

In order to determine the cosine value of y, use the fundamental identity:
$$\cos^{2}y+\sin^{2}y=1.$$
Knowing that the cosine value has to be negative, you should be able to figure out the answer.
2. 81,82,83:
Knowing the summation, double angle, half-angle formulae for the trigonometric functions should now give you the answers.
3.
85:
$$\sin^{2}\theta=\frac{1-\cos(2\theta)}{2}$$
4.86:
$$\sin(2\phi)=2\sin\phi\cos\phi, \sin\phi=\pm\sqrt{1-\cos^{2}\phi}$$

 

[tacman]

Hi there,

I understand your frustration and I'm happy to help you with these problems. Let's go through each one step by step:

81. To find sin(x-y), we can use the following formula: sin(x-y) = sin(x)cos(y) - cos(x)sin(y). Substituting the given values, we get: sin(x-y) = (5/13)(-4/5) - (12/13)(3/5) = -20/65 - 36/65 = -56/65.

83. To find sin 2x, we can use the double angle formula: sin 2x = 2sin(x)cos(x). Substituting the given values, we get: sin 2x = 2(5/13)(12/13) = 120/169.

84. To find sin x/2, we can use the half angle formula: sin x/2 = +/- √[(1-cosx)/2]. Since y is in Quadrant II, we know that cos(x) = -12/13. Substituting this into the formula, we get: sin x/2 = +/- √[(1-(-12/13))/2] = +/- √[(1+12/13)/2] = +/- √[25/26] = +/- 5/√26.

85. To find cos2(pheta), we can use the double angle formula: cos2(pheta) = cos^2(pheta) - sin^2(pheta). Substituting the given value, we get: cos2(pheta) = (2/3)^2 - (2/3)^2 = 4/9 - 4/9 = 0.

86. To find sin(2cos^-1(1/4)), we can use the inverse trigonometric function formula: sin(2cos^-1(x)) = 2√(1-x^2). Substituting the given value, we get: sin(2cos^-1(1/4)) = 2√(1-(1/4)^2) = 2√(1-1/16) = 2√(15/16) = √15/2.

I hope this helps you with your review and good luck on your exam tomorrow! Remember to always check your work and