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?

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 .

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2024年6月10日 · Click here 👆 to get an answer to your question ️ sin 75° + cos 65° ను 0° మరియు 45° మధ్యగల విలువల త్రికోణమి

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

2020年12月8日 · Find an answer to your question 7 Given that tan 25º = a, express each of the following in terms of a.a tan 205b sin 25°С cos 65° HAMZAWAHAJ HAMZAWAHAJ 08.12.2020

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

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2024年12月31日 · Step-by-step explanation: sin 25° cos 65° + cos 25° sin 65° Comparing with the identity: Sin( A + B ) = (Sin A × Cos B + Cos A × Sin B)