DSAT trig problem #sat #trigonometry

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another approach is saying sin(4k-22)=cos(6k-13)
remember cosx = sqrt(1-sin^2(x))
so then 2 both sides
so sin^2(4k-22)=1-sin^2(6k-13)
add sin^2(6k-13) to both sides and then
sin^2(4k-22)+sin^2(6k-13)=1
and the squares of sines and cosines only add up to 1 when the angles add up to equal 90 so we have
4k-22+(6k-13)=90
10k-35=90
10k=125
k=12.5

terrariariley
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help, I don’t quite understand what you meant from 0:19 - 0:30 😓

did you mean that b = 90 - a ?

i am so confused with this part but I understand everything else

mrinas
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Here's another approach..

By complementary angles .
Sin@=cos(90-@)
sin(4k-22)=cos(90-(4k-22)
cos(90-(4k-22)= cos(6k-13)
Cos literally cancels out
90-(4k-22)=6k-13
K=12.5..


Nice piece
.I love the background music 😂😄😄😄😄

danieldanmola
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Katya we can do the question without sin a⁰= cos b⁰
Can we?

qanoothamdani