Asked by Anonymous
I have to evaluate the following:
tan((sin^-1)(2/7))
and
sin^-1 (sin(17pi/6))
Please help.
tan((sin^-1)(2/7))
and
sin^-1 (sin(17pi/6))
Please help.
Answers
Answered by
Reiny
sin^-1 (2/7) means,
find the angle Ø , so that sinØ = 2/7
sketch a right-angled triangle with base angle Ø,
opposite side 2 and hyptenuse 7
by Pythagoras, adj^2 + 2^2 = 7^2
adjacent = √45 = 3√5
then tanØ = 2/(3√5)
thus tan(sin^-1 (2/7) = 2/(3√5)
for the second, I want you to really think about what you are looking at.
If you don't see it, follow my procedure.
find the angle Ø , so that sinØ = 2/7
sketch a right-angled triangle with base angle Ø,
opposite side 2 and hyptenuse 7
by Pythagoras, adj^2 + 2^2 = 7^2
adjacent = √45 = 3√5
then tanØ = 2/(3√5)
thus tan(sin^-1 (2/7) = 2/(3√5)
for the second, I want you to really think about what you are looking at.
If you don't see it, follow my procedure.
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