Question

3sin(2a-π/4)+2cos(3a-π) если a=π/4

Answer 1

$a=\frac{\pi}{4}\\\\3sin(2a-\frac{\pi}{4})+2cos(3a-\pi )=3sin(\frac{\pi }{2}-\frac{\pi}{4})+2cos(\frac{3\pi }{4}-\pi )=\\\\=3sin\frac{\pi }{4}+2cos(-\frac{\pi}{4})=3\cdot \frac{\sqrt2}{2}+2\cdot \frac{\sqrt2}{2}=\frac{5\sqrt2}{2}$

Answer 2

To evaluate the expression 3sin(2a-π/4) + 2cos(3a-π) when a = π/4, we'll substitute the value of 'a' and perform the calculations step by step.

Given: a = π/4

Now, let's calculate the values of sin(2a-π/4) and cos(3a-π) when a = π/4:

1. sin(2a-π/4): Substitute a = π/4 into the expression: sin(2(π/4) - π/4) = sin(π/2 - π/4) = sin(π/4) The sine of π/4 is 1/√2 or approximately 0.7071.

2. cos(3a-π): Substitute a = π/4 into the expression: cos(3(π/4) - π) = cos(3π/4 - π) = cos(-π/4) The cosine of -π/4 is the same as the cosine of π/4, which is 1/√2 or approximately 0.7071.

Now, we can substitute these values back into the original expression:

3sin(2a-π/4) + 2cos(3a-π) = 3(0.7071) + 2(0.7071) = 2.1213 + 1.4142 ≈ 3.5355

So, when a = π/4, the value of the expression 3sin(2a-π/4) + 2cos(3a-π) is approximately 3.5355.