2018年8月10日 · sin^2A+sin^2(A-B)+2sinAcosBsin(B-A)= sin^2a+(sinacosb-cosasinb)(sinacosb-cosasinb)+2sinacosb(sinbcosa-cosbsina)= sin^2a+sin^2acos^2b-2sinacosasinbcosb+cos^2asin^2b+2sinasinbcosacosb-2cos^2bsin^2a= sin^2a-sin^2acos^2b+cos^2asin^2b= sin^2a(1-cos^2b+cot^2asin^2b)= sin^2asin^2b(1+cot^2a)= sin^2asin^2b(csc^2a)= sin^2asin^2b(1/sin^2a)= sin^2b

2016年7月9日 · If sinA=4/5 and cosB= -5/13, where A belongs to QI and B belongs to QIII, then find sin(A+B).... See all questions in Sum and Difference Identities Impact of this question

2018年6月21日 · #"consider the right triangle with sides x and y and "# #"hypotenuse r, a is the angle between x and r"# #sina=y/r" and "cosa=x/r#

2019年3月14日 · 2sinA means that 2 times the value of sin A. However, it's totally different from sin 2A. Sin 2A Means that the value of x is doubled that is sin of two times x. Let's take x = 45° So, 2 sin 45 = 2*1/√2 = √2. And, sin 2A = sin 90° = 1. So you can clearly see that 2 sin A is not equal to sin 2A. sin 2A = 2 sin A cos A

2018年2月28日 · See below. I assume this is supposed to be: sin^2A+cos^2A=1 This is known as the Pythagorean Identity. This means it is true for all AinRR

Based on the Pythagorean Theorem we know #color(white)("XXX")sin^2(A)+cos^2(A)=1# or, more applicable in this case:

2018年3月19日 · We know that, #color(blue)((1) .x^4+y^4=(x^2+y^2)^2-2x^2y^2)# #color(red)((2) .x^6+y^6=(x^2+y^2)^3-3x^2y^2(x^2+y^2))#

2016年6月25日 · This is not an equation to be solved. The 'key word' here is #color(blue)"Prove"#.There are 3 options available to us here in attempting to prove this.

2014年11月6日 · by the trig identity #cos^2theta+sin^2theta=1#, #=pm sqrt{1-sin^2A}# by #sinA=1/sqrt{2}#,

2015年7月5日 · How do you prove #Cos^2 2A-cos^2 A=sin^2A-sin^2 2A#? Trigonometry Trigonometric Identities and Equations Proving Identities. 1 Answer