Answer:
-2sin(x) * sin(2x)
or any equivalent form, such as 4cos(x)[cos²(x)-1]
Step-by-step explanation:
simplify cos(3x) first:
cos(3x) = cos(2x+x) = cos(2x)cos(x) -sin(2x)sin(x)
using trig identities
= [2cos²(x)-1]cos(x) - [2sin(x)cos(x)]sin(x)
= 2cos³(x) - cos(x) - 2sin²(x)cos(x)
substituting using trig identity sin²(x) + cos²(x) = 1
2cos³(x) - cos(x) - 2[1-cos²(x)]cos(x)
2cos³(x) - cos(x) - 2cos(x)+2cos³(x)
4cos³(x) - 3cos(x)
remember this cos(3x), we still have to subtract cos(x)
4cos³(x) - 3cos(x) - cos(x) = 4cos³(x) - 4cos(x)
we can factor 4cos(x) to write this as a product of:
4cos(x)[cos²(x)-1]
further simplification if you want
trig identity sin²(x) + cos²(x) = 1
simplifying: sin²(x) = 1-cos²(x)
simplifying: -sin²(x) = cos²(x)-1
4cos(x)[cos²(x)-1]
4cos(x)[-sin²(x)]
-4cos(x)sin²(x)
trig identity: sin(2a) = 2cos(a)sin(a)
-2sin(x) * 2cos(x)sin(x)
-2sin(x)*sin(2x)
Answer:
1/5
Step-by-step explanation:
The original is 2/10 but simplified it is 1/5
Answer:
I believe it is jk the figure?
One of the roots of a quadratic equation is 7 - √3.
Because rational roots occur in conjugate pairs, the other root is 7 + √3.
The quadratic equation is
(x -(7-√3))(x - (7+√3))
= x² - (7+√3)x - (7-√3)x + (49-3)
= x² - 7x - (√3)x - 7x + (√3)x + 46
= x² - 14x + 46
Answer: x² - 14x + 46
Answer:
Part 1
Type II error
Part 2
No ; is not ; true
Step-by-step explanation:
Data provided in the question
Mean = 100
The Random sample is taken = 43 students
Based on the given information, the conclusion is as follows
Part 1
Since it is mentioned that the classes are successful which is same treated as a null rejection and at the same time it also accepts the alternate hypothesis
Based on this, it is a failure to deny or reject the false null that represents type II error
Part 2
And if the classes are not successful so we can make successful by making type I error and at the same time type II error is not possible
Therefore no type II error is not possible and when the null hypothesis is true the classes are not successful
