SOLUTION: Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5 (Both are in Quadrant II.) sec (v - u)

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Question 1060297: Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5 (Both are in Quadrant II.) sec (v - u)
Found 3 solutions by ikleyn, Edwin McCravy, MathTherapy:
Answer by ikleyn(52817) About Me  (Show Source):
You can put this solution on YOUR website!
.
Yesterday I solved very similar problem under the link
https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1060117.html

https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1060117.html

Read it attentively. It is your sample.
Then solve your problem applying the same method and technique.


Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

sec%28v+-+u%29%22%22=%22%221%2Fcos%28v-u%29%22%22=%22%221%2F%28cos%28v%29cos%28u%29%2Bsin%28v%29sin%28u%29%29%22%22=%22%221%5B%22%22%5D%2F%28%28-3%2F5%29cos%28u%29%2Bsin%28v%29%285%2F13%29%29%22%22=%22%22

We have to pause here and use an identity:

sin%5E2%28theta%29%2Bcos%5E2%28theta%29%22%22=%22%221

we use it to find cos(u) from sin(u):

sin%5E2%28u%29%2Bcos%5E2%28u%29%22%22=%22%221

%285%2F13%29%5E2%2Bcos%5E2%28u%29%22%22=%22%221

25%2F169%2Bcos%5E2%28u%29%22%22=%22%221

Multiply through by 169

25%2B169cos%5E2%28u%29%22%22=%22%22169

Subtract 25 from both sides:

169cos%5E2%28u%29%22%22=%22%22144

cos%5E2%28u%29%22%22=%22%22144%2F169

cos%28u%29%22%22=%22%22%22%22+%2B-+sqrt%28144%2F169%29

cos%28u%29%22%22=%22%22%22%22+%2B-+12%2F13

We can determine which sign + or - to use because
we know that both are in Quadrant II, and the
cosine is negative in quadrant II.  So cos%28u%29%22%22=%22%22-12%2F13

we can now substitute that in 

1%5B%22%22%5D%2F%28%28-3%2F5%29cos%28u%29%2Bsin%28v%29%285%2F13%29%29%22%22=%22%22

1%5B%22%22%5D%2F%28%28-3%2F5%29%28-12%2F13%29%2Bsin%28v%29%285%2F13%29%29%22%22=%22%22

But we still need sin(v).  So we go back to the identity:

sin%5E2%28theta%29%2Bcos%5E2%28theta%29%22%22=%22%221

we use it to find sin(v) from cos(v):

sin%5E2%28v%29%2Bcos%5E2%28v%29%22%22=%22%221

sin%5E2%28v%29%2B%28-3%2F5%29%5E2%22%22=%22%221  

sin%5E2%28v%29%2B9%2F25%22%22=%22%221

Multiply through by 25

25sin%5E2%28v%29%2B9%22%22=%22%2225

Subtract 9 from both sides:

25sin%5E2%28v%29%22%22=%22%2216

sin%5E2%28v%29%22%22=%22%2216%2F25

sin%28v%29%22%22=%22%22%22%22+%2B-+sqrt%2816%2F25%29

sin%28v%29%22%22=%22%22%22%22+%2B-+4%2F5

As before we can determine which sign + or - to use because
we know that both are in Quadrant II, and the sine is
positive in quadrant II.  So sin%28v%29%22%22=%22%22%22%22+%2B+4%2F5

we can now substitute that in 


1%5B%22%22%5D%2F%28%28-3%2F5%29%28-12%2F13%29%2B%284%2F5%29%285%2F13%29%29%22%22=%22%221%5B%22%22%5D%2F%28%2836%2F65%29%2B%2820%2F65%29%29%22%22=%22%22%281%5B%22%22%5D%29%2F%2856%2F65%29%22%22=%22%221%22%F7%2256%2F65%22%22=%22%221%22%22%2A%22%2265%2F56%22%22=%22%2265%2F56

Answer: 65%2F56

Edwin

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
Find the exact value of the trigonometric function given that sin u = 5/13 and cos v = -3/5 (Both are in Quadrant II.) sec (v - u)
sin+%28u%29+=+5%2F13
This is a 5-12-13 Pythag triple, so: cos+%28u%29+=+-+12%2F13 --- cos is < 0 in the 2nd quadrant
cos+%28v%29+=+%28-+3%29%2F5
This is a 3-4-5 Pythag triple, so: sin+%28v%29+=+4%2F5 -------- sin is > 0 in the 2nd quadrant
====> 1%2F%28%28-+3%2F5%29+%2A+%28-+12%2F13%29+%2B+%284%2F5%29+%2A+%285%2F13%29%29 ======> 1%2F%2836%2F65+%2B+20%2F65%29 ====> highlight_green%281%2F%2856%2F65%29+=+65%2F56%29