i have two problems that i can't figure out how to do... 1. Give the exact real number value for Tan inverse [tan 7pi/12] and 2. Evaluate: (show your work. do not give a decimal approximation) Sin[cos inverse(1/3) + sin inverse(-2/5)]
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Originally Posted by ilaura i have two problems that i can't figure out how to do... 1. Give the exact real number value for Tan inverse [tan 7pi/12] and 2. Evaluate: (show your work. do not give a decimal approximation) Sin[cos inverse(1/3) + sin inverse(-2/5)] On a restriced interval is an inverse function of so and for the second one use the fact that and after using the identity
i'm still really confused..... i tried doing something with the second one, i don't think its right but i got: 6(square root of 2) + 3(square root of 21) -1 ____________________ 15
Originally Posted by ilaura i'm still really confused..... i tried doing something with the second one, i don't think its right but i got: 6(square root of 2) + 3(square root of 21) -1 ____________________ 15
Originally Posted by Mathstud28 ......and for the second one use the fact that and i don't understand where these come from.... are they some kind of identities?
Originally Posted by ilaura i don't understand where these come from.... are they some kind of identities? Yes List of trigonometric identities - Wikipedia, the free encyclopedia
Originally Posted by ilaura [snip] and 2. Evaluate: (show your work. do not give a decimal approximation) Sin[cos inverse(1/3) + sin inverse(-2/5)] Let . Let . Then
Originally Posted by Mathstud28 where did you get the 1/2? is it part of the identity?
Originally Posted by ilaura where did you get the 1/2? is it part of the identity? Clearly it's a typo and should be 1/3.
Originally Posted by ilaura My homework says: 1. Give the exact real number value for arctan [tan (7pi/12)] steps would be appreciated. thanks Note that arctangent and tangent are inverses of each other. Thus, if we have , it just reduces to . The same thing applies to . It also equals . Knowing this, what do you think the answer is?
Last edited by Chris L T521; Nov 30th 2008 at 12:22 PM.
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