I'm bigger than the Log of Planck constant!
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1
0
9
R
e
(
tan
(
2
+
9
i
)
)
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Evaluate the expression above.
Details and Assumptions :
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i = − 1
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Always use radian values unless stated otherwise.
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You may use a calculator solve this problem.
The answer is -24.
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1 solution
Okay , I'll remember that such an easy question like this touched the 300 point mark ! (305 actually)
EDIT : 310
EDIT 2 : 335
EDIT 3 : 380 !
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The reason is 2% rule. Approximation on B'ant is valid for ±2% of answer. Many people answered 0 which is incorrect as answer itself is too small
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Oh , thanks ! I didn't know of this fact :)
Actually , the digital sum of my JEE roll number is 29 , so that's the inspiration for this question !
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Feeling pressure for JEE? Last question? Really? That's sad! But joy and happiness are coming shortly in your life! I hope for the best!
What identity is that in the third to last line?
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First of all I had multiplied and divided the term cos ( 2 − 9 i ) .
Now I have used the following identities :
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sin ( A + B ) + sin ( A − B ) = 2 sin A cos B
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cos ( A + B ) + cos ( A − B ) = 2 cos A cos B
Ok so we have to find the real part of tan ( 2 + 9 i )
tan ( 2 + 9 i ) = cos ( 2 + 9 i ) sin ( 2 + 9 i ) = 2 cos ( 2 + 9 i ) cos ( 2 − 9 i ) 2 sin ( 2 + 9 i ) cos ( 2 − 9 i ) = cos 4 + cos ( 1 8 i ) sin 4 + sin ( 1 8 i ) = cos 4 + cosh ( 1 8 ) sin 4 + i sinh ( 1 8 )
The rest is obvious I guess !