Ithen did the same thing to the other side to get $$-2(\sin(B+C)\cos(B+C)+\sin(A+C)\cos(A+C)+\sin(A+B)\cos(A+B))$$ and then tried using the compound angle formula to see if i got an equality. However the whole thing became one huge mess and I didn't seem to get any closer to the solution. Youhave (b+c)^2=(20-a)^2 and so b^2+c^2-bc=a^2-40a+400-120. But by the cosine formula that is a^2, so you have 40a=280 and hence a=7, so b+c=13,bc=40. Hence b,c=5,8. wecan draw the integral of sin (x), ∫f (x) dx = −cos (x) + C (C= constant of integration) and derivative of sin (x),f′ (x) as cos (x). We can determine the values of sine function as positive or negative depending upon the quadrants. Here is a table where we can see that on one hand sine 270 is negative and on the other hand sine 90 is Thearea of a triangle can be represented by the expression, One-half b c sine (60 degrees), StartFraction a squared b sine (60 degrees) Over 2 EndFraction, One-half a squared sine (60 degrees), the correct options are B,C, and D.. What is a Triangle? A triangle is a polygon with three sides, three angles and three vertices.. Triangle A B C is an equilateral triangle. Perimeterand Area Winds, Storms and Cyclones The Triangle and Its Properties. Verb. Click here👆to get an answer to your question ️ In any triangle ABC, prove that sin (B - C)sin (B + C) = b^2 - c^2a^2 . Explanationfor the correct option. Given that, sin A + sin B + sin C = 3 ( 1). Recall: sin 90 ο = 1. If we assume, A = B = C = 90 ο then the equation (1) is satisfied. So, substitute A = B = C = 90 ο in cos A + cos B + cos c we get: cos A + cos B + cos C = cos 90 ο + cos 90 ο + cos 90 ο = 0 + 0 + 0 = 0. Generalizedtrigonometry Reference Identities Exact constants Tables Unit circle Laws and theorems Sines Cosines Tangents Cotangents Pythagorean theorem Calculus Trigonometric substitution Integrals ( inverse functions) Derivatives v t e Sothen I tried, $\sin\left(\frac{-B-C + \pi}{2}\right)\cdots$ and this didn't go anywhere either. I don't know what to try, and I've seen other people's solutions and they do something like: $\sin(C) = \sin(A + B)$, $\cos(C/2) = \sin(\frac{A + B}{2})$ but i don't see where they got this part from. FU92Z.