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
B. RSP and PST I’m pretty sure.
Answer:
z=9.7
Step-by-step explanation:
3z + 8 - 2 = -35
Subtract 6 from both sides
3z=29
divide 3 from both sides
z=9.7
Answer:
15
Step-by-step explanation:
Use the pythagorean theorem.
A² + B² = C²
In this case, A = 8, B is unknown, and C = 17. B is the side that we want to calculate.
8² + B² = 17²
64 + B² = 289
B² = 225
B = √225 = 15
