2016年2月18日 · Click here 👆 to get an answer to your question ️ is it right to say that sin(A+B)=sinA+sinB gurika0chjotki gurika0chjotki 19.02.2016

2011年8月6日 · I am sorry, I just don't follow it that well. Why does there HAVE to be a point C whose gradient is equal to the gradient of the secant AB?

2018年8月24日 · Then find value of SinA+SinB if A + B = 90° is √2. Step-by-step explanation: In order to solve the above question, we need to know some of the trigonometric formulae:-According to the question, A + B = 90° and we need to find the value of sinA+ sinB. From A + B = 90' we can evaluate the value of B in terms of A i.e. B = 90° - A. Let x ...

2016年4月26日 · Prove that sinA·sinB <= sin^2(C/2) Discuss the case of equality. My attempt : I used Apollonius' Theorem to get -> a^2 + b^2 = 2(BD)c {where a, b, c are sides opposite to angle A, B, C respectively in triangle ABC.} and tried to get the result using the triangle inequalities and Sine Law but couldn't figure out the way to use them to get the ...

2021年8月8日 · Sin(A+B) =SinA+SinB. To prove : LHS = RHS. Solution : As we know , ⇒ sin(A+B) = sinAcosB+sinBcosA. So,if cos function can be eliminated from RHS,then we will get the required result. To eliminate cos function,let us take some cases for values of A and B. For RHS. When A is 0° sin0°×cosB+sinB×cos0° 0 + sinB × 1. sinB. When A is 90°

2019年4月14日 · We have to prove : cosa cos(b-a)-sina sin(b-a) = cosb. Trigonometric formulas to be used : → cos(A-B) = cosAcosB + sinAsinB. → sin(A-B) = sinAcosB-cosAsinB. Now using those to solve the LHS : = cosa [cosbcosa + sinbsina] - sina [sinbcosa - cosbsina] = (cosa × cosb cosa + cosa × sina sinb) - (sina × sinb cosa - sina × cosb sina)

2018年12月23日 · Find an answer to your question Sina+sinb+sinc=0 cosa+cosb+cosc=0 by complex numbers shivramnahak3948 shivramnahak3948 23.12.2018

SOLUTION:- . ⇒Given parameters: sinA = 1. sinB = 2. sinC = 3. b = 4 cm. ⇒According to the sin rule. ⇒(a/sin A) = (b/sin B) = (c/sin C)

2020年5月31日 · (sinA + sinB / cosA - cosB)^2015 So putting the values in the expression, we get: (2 sin(a+b/2) cos(a-b/2))^2015 / (-2 sin(a+b/2) sin(a-b/2))^2015

2017年10月15日 · Answer:[tex] sinA+sinB-sinC=4sin\frac{A}{2} sin\frac{B}{2} cos\frac{C}{2} [/tex]Step-by-step explanation:[tex]LHS = sinA+sinB-sinC[/tex][tex] = 2sin\left(\frac{…