so, use the cos formula, instead of sin. What's the trouble?
cos(α−β) = cosα cosβ + sinα sinβ
You have all the values; just plug 'em in.
Hi Steve and Damon, if you're reading this, please take a look at my previous post because I had to apologize for my mistake.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ My Home Work Question:
Suppose that α is an angle such that tanα=5/7 and sinα<0. Also, suppose that β is an angle such that cotβ=−3/5 and cscβ>0. Find the exact value of sine(α−β).
3 answers
As Steve and Damon pointed out to you earlier,
sin(α−β) = sinαcosβ - cosαsinβ
given: tanα=5/7 and sinα<0 , so by the CAST rule
α must be in quadrant III
since tanα = 5/7, by Pythagoras, the hypotenuse of our triangle is √(25+49) = √74
thus sinα = -5/√74 and cosα = -7/√74
similarily, if cotβ=−3/5 and cscβ>0 , β must be in quadrant II, and tanβ = -5/3
sinβ = 5/√34 , and cosβ = -3/√34
You now have the exact 4 values needed to sub into my formula in the second line.
sin(α−β) = sinαcosβ - cosαsinβ
given: tanα=5/7 and sinα<0 , so by the CAST rule
α must be in quadrant III
since tanα = 5/7, by Pythagoras, the hypotenuse of our triangle is √(25+49) = √74
thus sinα = -5/√74 and cosα = -7/√74
similarily, if cotβ=−3/5 and cscβ>0 , β must be in quadrant II, and tanβ = -5/3
sinβ = 5/√34 , and cosβ = -3/√34
You now have the exact 4 values needed to sub into my formula in the second line.
THANK YOU GUYS SOOO MUCH, GOD BLESS.
3 answers