To the 1000000th power of 5

To the 1000000th power of 5

x=5^1000000
lgx=1000000lg5 ≈ 698970
x ≈ 10^698970
That is, 5 ^ 1000000 is about to add 698970 zeros after 10

What are the last two digits of the one million power of 1976

Actually, that's just the calculation of the last two digits, 76 ^ 1 = 76 ^ 2 = 5776 ^ 3 = 438976
So the answer is 76, no matter what

How does 4sin square (B - π / 4) become 2cos (2B - π / 2) + 2

First of all, we should know the formula Cos2 α = 1-2sin & # 178; α
We obtain that 2Sin & # 178; α = 1-cos2 α
∴4sin²（b-π/4)=2*（1-cos（2b-π/2））=2-2cos（2b-π/2）
There is something wrong with your answer
Hope to help you, if you have any questions, please ask o (∩)_ (laughter)~

Given Tan α = 3, find 3 / 4sin square α + 1 / 2cos square α =?

tanα=3
∴ sinα/cosα=3
That is sin α = 3cos α
Substituting Sin & # 178; α + cos & # 178; α = 1
∴ 9cos²α+cos²α=1
∴ cos²α=1/10,∴ sin²α=9/10
3 / 4sin square α + 1 / 2cos square α
=(3/4)*(9/10)+(1/2)*(1/10)
=27/40+2/40
=29/40

Calculation: 2cos (- 660 °) - sin630 ° / 4cos1020 ° + 2cos (- 300 °)

[2cos(-660°）-sin630°]/[4cos1020°+2cos（-300°）]
=[2cos(-720+60)-sin(720-90)]/[4cos(720+300)+2cos300]
=(2cos60+sin90)/(4coos300+2cos33)
=(2cos60+1)/6cos60
=2/3
sinα=sin(±360n+α)
sin(-a)=-sina
cos(-a)=cosa
sin90=1

How to calculate if θ is 30 ° in 2cos ^ 2 θ - 1?
Do you want to replace theta with 30 ° and calculate it by 2cos60 ° - 1? Or is there any other formula? What do you need to pay attention to when doing this kind of problem? When theta equals 60 °, how to do cos ^ 2 θ sin ^ 2 θ? Explain o (∩)_ Thank you~~

Excuse me, do you want to express cos 2 θ or cos ^ 2 θ? These two are completely different. The latter one represents the quadratic power of COS!

The value of 2cos & # 178; 22.5 ° - 1 is

2cos² 22.5 °-1=cos45°=√2/2

2cos²22.5º-1

2cos²22.5º-1=cos45°=√2/2

Sequence 1,2cos θ, (2cos θ) square, (2cos θ) cubic The sum of the first 100 terms is 0 θ∈ (0180 degrees) to find θ

Let me have a try
a1=1,q=2cosθ
So an = q ^ (n-1)
Obviously Q ≠ 0,1, otherwise S100 > 0
Sn=(1-q^n)/(1-q)
S100=(1-q^100)/(1-q)=1+q+q^2+...+q^99=0
q+q^2+...+q^99=q(1+q+q^2+...+q^98)=qS99=-1
S99=-1/q=S100-a100=-a100
So A100 = 1 / Q = q ^ 99
q^100=1,q≠1
So q = 2cos, θ = - 1, θ∈ (0180)
The solution is θ = 120

If α is an acute angle and 2cos ^ 2 + 7SiN α - 5 = 0, find the degree of α

2cos²α+7sinα-5=0
2(1-sin²α)+7sinα-5=0
2sin²α-7sinα+3=0
sinα=[7±√(7^2-4X2X3)/(2X2)
=(7±5)/4
That is: sin α = 3
sinα=1/2
Because α is an acute angle
α=30°