If you are doing Taylor expansions, perhaps just work out the Taylor expansion for cos(65). Alternately, since 65 = 60 + 5, you could use the cosine addition formula to work out cos(60 + 5). Since 5 is pretty small, you could make an approximation to simplify your work for cos(5).

2016年3月31日 · Solve, in the interval $0^\circ \leq \theta \leq 360^\circ$, $$\cos(\theta + 25^\circ) + \sin(\theta +65^\circ) = 1$$ I've attempted expanding but reach a point with no common factors & see how to manipulate the trig ratios to move on; is there a solution without expanding?

2019年3月26日 · Prove that cos pi/65.cos 2pi/65.cos 4pi/65.cos 8pi/65.cos 16pi/65.cos 32 pi/65 = 1/64? - 9017381 deeptisomasekar3933 deeptisomasekar3933 26.03.2019

2019年6月9日 · Answer: Sin 25° cos 65° + cos25° sin 65° Let 25°. = A. 65° = B. It is in the form Sin A cos B + cos A sin B ...

2021年1月18日 · ANSWERsin65 o +cos65 o =sin(45 o +20 o )+cos(45 o +20 o )=sin45 o cos20 o +cos45 o sin20 o +cos45 o cos20 o −sin45 o sin20 o = 2 cos20 Advertisement New questions in Math

Therefore, the $\sin(85)$'s cancel and you're left with $$ \sin(20)\cos(65)-\sin(65)\cos(20)=-\sin(45). $$ By evaluating this directly, you have the value you're looking for. Share Cite

2015年2月17日 · $65^\circ$ isn't a nice one, unfortunately: the only constructible angles with natural degree measure are multiples of $3^\circ$. This is a little complicated to prove directly - a problem since antiquity, the last piece of the puzzle didn't appear until 1837, when Wantzel proved that $20^\circ$ was unconstructible .

2018年7月1日 · Given x = cos 55, y = cos 65, z = cos 175. We need to find xy, yz, zx. = > cos 55 * cos 65 + cos 65 * cos 175 + cos 175 * cos 55

2023年2月8日 · The expression Sin 65° cos 25 + Sin 25 cos 65° is equal to 1 + Sin 40°. Advertisement tanishpatel0512 tanishpatel0512

2022年1月5日 · Thus, answer is sin 75° + cos 65° = cos 15° + sin 25°. Advertisement anakapallterapathrao anakapallterapathrao