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sin2A ≡ 2sinAcosA. cos2A ≡ cos2 A − sin2 A ... For an expression for sin3A, write it as sin(2A + A) then use both compound and double angle formulae. May 21, 2020 ... If sinA=(3/5) and A is in first quadrant, then the values of sin2A, cos2A and tan2A are (A) 24/25, 7/25, 24/7 (B) 1/25, 7/25, 1/7 (C) 24/25, ... Jan 25, 2021 ... I got an answer but not sure if it's correct: sin2A = 2sinAcosA (double angle formula) =2sinA(y/2) (given at start) =(2ysinA)/2 =ysinA What does it mean to prove a trigonometric identity? How do you prove cscθ×tanθ=secθ? How do you prove (1−cos2x)(1+cot2x)=1? How do you show that 2sinxcosx=sin2 ... ... sin2a=NULL, *cos2a=NULL, *sincosa=NULL; static double elev, sinl, cosl ... (sin2a != NULL) free(sin2a); if(cos2a != NULL) free(cos2a); if(sincosa ... Substituting this in the above formula, sin2A = 2 sin A √(1 - sin2A). This formula is in the terms of sin or sine function only. Oct 18, 2019 ... Sin4A+sin2A divide by 1+cos2A+cos4A= tan2A prove it · Expert-Verified Answer · Answer · New questions in Math. Construct a triangle with sides 5 ... Apr 8, 2025 ... ... sin2a α + cos² 1 · May be an image of text that says 'TRIGONOMETRIC FUNCTIONS FOR A RIGHT TRIANGLE a · May be a graphic of text that says ... Comparing with the given expression, we get ⇒ a = 1, b = 3, c = 3 ⇒ a + b + c = 1 + 3 + 3 = 7 Therefore, the required value of a + b + c is 7. sin2A+cos2A=1,tan2A+ ... sin2A=2cos2A−1=1−2sin2 ...