2 cosy+siny =?
could you be a bit more specific?
one possibility:
2cosy + siny = √5(2/√5 cosy + 1/√5 siny)
so, if sinx = 2/√5, that is just
√5 sin(y+x)
Well, if you're looking for a mathematical answer, the sum of 2cos(y) + sin(y) would be the value you get when you substitute a specific value for 'y'. However, if you're looking for a more humorous answer, I could say that 2cos(y) + sin(y) equals the hilarious sound of a confused owl trying to do math while juggling feathers.
To simplify the expression 2cos(x) + sin(x), we can apply trigonometric identities. Here's the step-by-step solution:
Step 1: Recognize the sum-to-product identity for cosine: cos(a) + cos(b) = 2cos((a+b)/2)cos((a-b)/2)
Here, let a = x and b = -(π/2), giving us:
cos(x) + cos(-π/2) = 2cos((x - π/2)/2)cos((x + π/2)/2)
Step 2: Simplify cos(-π/2) using the unit circle.
The unit circle shows that cos(-π/2) = 0. Thus, we have:
cos(x) + 0 = 2cos((x - π/2)/2)cos((x + π/2)/2)
Step 3: Simplify the expression by applying the half-angle identity for cosine: cos(θ/2) = ±√[(1 + cos(θ))/2]
Therefore, we have:
2cos((x - π/2)/2)cos((x + π/2)/2) =
2[cos(x/2 - π/4)cos(x/2 + π/4)] =
2[cos(x/2)cos(π/4) - sin(x/2)sin(π/4)] =
√2[cos(x/2) - sin(x/2)]
Step 4: Substitute the simplified expression back into the original expression: 2cos(x) + sin(x) =
√2[cos(x/2) - sin(x/2)]
So, 2cos(x) + sin(x) simplifies to √2[cos(x/2) - sin(x/2)].
To find the value of 2cos(y) + sin(y), we'll need to use the properties of trigonometric functions and some algebra.
First, let's rewrite the expression using the identities:
2cos(y) + sin(y) = 2cos(y) + 1*sin(y)
Now, we can use the angle addition formula for cosine:
cos(x + y) = cos(x)cos(y) - sin(x)sin(y)
Let's assume x = 0:
cos(y) = cos(0 + y) = cos(0)cos(y) - sin(0)sin(y)
= 1*cos(y) - 0*sin(y)
= cos(y)
Now, we can substitute this back into our original expression:
2cos(y) + sin(y) = 2cos(y) + 1*sin(y)
= 2*cos(y) + sin(y)
= 2*cos(y) + cos(y) (by substituting cos(y) for cos(0 + y))
= 3*cos(y)
So, 2cos(y) + sin(y) simplifies to 3cos(y).