Answer:
A, C
Step-by-step explanation:
Actually, those questions require us to develop those equations to derive into trigonometrical equations so that we can unveil them or not. Doing it only two alternatives, the other ones will not result in Trigonometrical Identities.
Examining
A) True
Double angle 
B) False,
No further development towards a Trig Identity
C) True
Double Angle Sine Formula 
D) False No further development towards a Trig Identity
![[sin(x)-cos(x)]^{2} =1+sin(2x)\\ sin^{2} (x)-2sin(x)cos(x)+cos^{2}x=1+2sinxcosx\\ \\sin^{2} (x)+cos^{2}x=1+4sin(x)cos(x)](https://tex.z-dn.net/?f=%5Bsin%28x%29-cos%28x%29%5D%5E%7B2%7D%20%3D1%2Bsin%282x%29%5C%5C%20sin%5E%7B2%7D%20%28x%29-2sin%28x%29cos%28x%29%2Bcos%5E%7B2%7Dx%3D1%2B2sinxcosx%5C%5C%20%5C%5Csin%5E%7B2%7D%20%28x%29%2Bcos%5E%7B2%7Dx%3D1%2B4sin%28x%29cos%28x%29)
[tex]\begin{gathered} \text{The inequality is,} \\ 5
In mathematics there is a rule of exponents where we can "distribute" the powers/exponents in the numerator and denominator of any expression. Therefore, given an expression as 
, the exponent n can be "distributed" as: 
.
In our case, the power, n=11; a=7 and b=4. Thus, the expression 
, can be written as 
.
Thus, out of the given options, option A is the correct option.
Hello from MrBillDoesMath!
Answer:
See Discussion below
Discussion:
(sinq + cosq)^2 = => (a +b)^2 = a^2 + 2ab + b^2
(sinq)^2 + (cosq)^2 + 2 sinq* cosq => as (sinx)^2 + (cosx)^2 = 1
1 + 2 sinq*cosq (*)
Setting a = b = q in the trig identity:
sin(a+b) = sina*cosb + cosa*sinb
sin(2q) = (**)
sinq*cosq + cosq*sinq => as both terms are identical
2 sinq*cosq
Combining (*) and (**)
(sinq + cosq)^2 = 1 + 2sinq*cosq => (**) 2sinq*cosq = sqin(2q)
= 1 + sin(2q)
Hence
(sinq + cosq)^2 = 1 + sin(2q) => subtracting 1 from both sides
(sinq + cosq)^2 - 1 = sin(2q)
The last statement is what we are trying to prove.
Thank you,
MrB
Answer:
the answer is D becuase it is the only one with a range of -18
