The main identity you need is the double angle one for cosine:
![\cos^2x=\dfrac{1+\cos2x}2](https://tex.z-dn.net/?f=%5Ccos%5E2x%3D%5Cdfrac%7B1%2B%5Ccos2x%7D2)
We get
![\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8](https://tex.z-dn.net/?f=%5Ccos%5E6x%3D%28%5Ccos%5E2x%29%5E3%3D%5Cleft%28%5Cdfrac%7B1%2B%5Ccos2x%7D2%5Cright%29%5E3%3D%5Cdfrac%7B%281%2B%5Ccos2x%29%5E3%7D8)
Expand the numerator to apply the identity again:
![\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8](https://tex.z-dn.net/?f=%5Ccos%5E6x%3D%5Cdfrac%7B1%2B3%5Ccos2x%2B3%5Ccos%5E22x%2B%5Ccos%5E32x%7D8)
![\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8](https://tex.z-dn.net/?f=%5Ccos%5E6x%3D%5Cdfrac%7B1%2B3%5Ccos2x%2B3%5Cleft%28%5Cfrac%7B1%2B%5Ccos2%282x%29%7D2%5Cright%29%2B%5Ccos2x%5Cleft%28%5Cfrac%7B1%2B%5Ccos2%282x%29%7D2%5Cright%29%7D8)
![\cos^6x=\dfrac{1+3\cos2x+\frac32+\frac32\cos4x+\frac12\cos2x(1+\cos4x)}8](https://tex.z-dn.net/?f=%5Ccos%5E6x%3D%5Cdfrac%7B1%2B3%5Ccos2x%2B%5Cfrac32%2B%5Cfrac32%5Ccos4x%2B%5Cfrac12%5Ccos2x%281%2B%5Ccos4x%29%7D8)
![\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{16}\cos2x\cos4x](https://tex.z-dn.net/?f=%5Ccos%5E6x%3D%5Cdfrac5%7B16%7D%2B%5Cdfrac7%7B16%7D%5Ccos2x%2B%5Cdfrac3%7B16%7D%5Ccos4x%2B%5Cdfrac1%7B16%7D%5Ccos2x%5Ccos4x)
Finally, make use of the product identity for cosine:
![\cos2x\cos4x=\dfrac{\cos6x+\cos2x}2](https://tex.z-dn.net/?f=%5Ccos2x%5Ccos4x%3D%5Cdfrac%7B%5Ccos6x%2B%5Ccos2x%7D2)
so that ultimately,
![\cos^6x=\dfrac5{16}+\dfrac7{16}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos2x+\dfrac1{32}\cos6x](https://tex.z-dn.net/?f=%5Ccos%5E6x%3D%5Cdfrac5%7B16%7D%2B%5Cdfrac7%7B16%7D%5Ccos2x%2B%5Cdfrac3%7B16%7D%5Ccos4x%2B%5Cdfrac1%7B32%7D%5Ccos2x%2B%5Cdfrac1%7B32%7D%5Ccos6x)
![\cos^6x=\dfrac5{16}+\dfrac{15}{32}\cos2x+\dfrac3{16}\cos4x+\dfrac1{32}\cos6x](https://tex.z-dn.net/?f=%5Ccos%5E6x%3D%5Cdfrac5%7B16%7D%2B%5Cdfrac%7B15%7D%7B32%7D%5Ccos2x%2B%5Cdfrac3%7B16%7D%5Ccos4x%2B%5Cdfrac1%7B32%7D%5Ccos6x)
Answer:
5/8x+1/2x-4=5
or, 5(2x-4)+8x= 5
or, 10x- 20+8x=5
or, 10x-8x= 20+5
or, 2x= 25
or,x= 25/2
:.x= 25/2
Step-by-step explanation:
first we have to make denominator then we can criss cross multiply. we can make in same point .if there is plus we can subtract and if there is front minus if we take it back it must be plus. then we can put same point then we can divide
Answer:
6; 9; 12
Step-by-step explanation:
It's possible because they cut cakes into different no. of slices
Luke: 2/3 of the cake are 4 slices
Entire cake 4/(2/3) = 4×3/2 = 6 slices
Kira: 2/3 of the cake are 6 slices
Entire cake 6/(2/3) = 6×3/2 = 9 slices
Ali: 2/3 of the cake are 8 slices
Entire cake 8/(2/3) = 8×3/2 = 12 slices
Answer:
23/50
Step-by-step explanation:
27 + 23 = 50 (Implying that 50 is the total amount of tickets).
23 of them were sold to adults, so basically 23/50.
Remember a ratio can be written in 3 forms, so it can be either of these:
23:50
23/50
23 to 50