Evaluation of limx→0sin⁡(x+a)−sin⁡(a)sin⁡2x I have got as far as below using sin⁡a+sin⁡b in numerator and sin⁡2a in the denominator but am not sure how to expand further, or if these are the right rules to apply. limx→0sin⁡(x)cos⁡(a)+sin⁡(a)[cos⁡(x)−1]2sin⁡(x)cos⁡(x) I need to simplify it down to apply limx→0sin⁡(x)x

Question

Evaluation of limx→0sin⁡(x+a)−sin⁡(a)sin⁡2x  I have got as far as below using sin⁡a+sin⁡b in numerator and sin⁡2a in the denominator but am not sure how to expand further, or if these are the right rules to apply.  limx→0sin⁡(x)cos⁡(a)+sin⁡(a)[cos⁡(x)−1]2sin⁡(x)cos⁡(x)  I need to simplify it down to apply limx→0sin⁡(x)x

Gwendolyn Meyer · Accepted Answer

Continue with this:  sin⁡xcos⁡a2sin⁡xcos⁡x+sin⁡a[cos⁡x−1]2sin⁡xcos⁡x  cos⁡a2cos⁡x−sin⁡a2sin2x24sin⁡x2cos⁡x2cos⁡x   cos⁡a2cos⁡x−sin⁡asin⁡x22cos⁡x2cos⁡x   then  limx→0cos⁡a2cos⁡x−sin⁡asin⁡x22cos⁡x2cos⁡x=cos⁡a2×1−0=cos⁡a2

ebbonxah · Accepted Answer

ddasin⁡a=limx→0sin⁡(x+a)−sin⁡ax=cos⁡a   limx→0sin⁡(x+a)−sin⁡asin⁡2x=limx→0(sin⁡(x+a)−sin⁡a)xxsin⁡2x=limx→0(sin⁡(x+a)−sin⁡ax)(xsin⁡2x)=12cos⁡a

Evaluation of \lim_{x \to 0} \frac{\sin(x+a) -\sin(a)}{\sin2x} I have got as far as below

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Answer & Explanation

New Questions in Precalculus