tan1'*tan2'*tan3'…… Tan 89 '

tan1'*tan2'*tan3'…… Tan 89 '

tan89=cot(90-89)=cot1=1/tan1
So TAN1 * tan89 = 1
In the same way
tan2*tan88=1
……
tan44*tan46=1
tan45=1
So the original formula = 1

cos(60+a)+sin(30+a) Use induction formula!

cos(60+a)+sin(30+a)
=cos60cosa-sin60sina+sin30cosa+cos30sina
=0.5cosa - (GEN.3) / 2sina + 0.5cosa + - (GEN.3) / 2sina
=cosa
Alas, you don't give a point, the conditions are so harsh, there are too many induction formulas, no matter which method you use or this way
Formula 1:
Let α be an arbitrary angle, and the values of the same trigonometric function of the angle with the same end edge are equal
sin（2kπ＋α）＝sinα
cos（2kπ＋α）＝cosα
tan（2kπ＋α）＝tanα
cot（2kπ＋α）＝cotα
Formula 2:
Let α be an arbitrary angle, the relationship between the trigonometric function value of π + α and the trigonometric function value of α
sin（π＋α）＝－sinα
cos（π＋α）＝－cosα
tan（π＋α）＝tanα
cot（π＋α）＝cotα
Formula 3:
The relationship between trigonometric function values of arbitrary angle α and - α is as follows
sin（－α）＝－sinα
cos（－α）＝cosα
tan（－α）＝－tanα
cot（－α）＝－cotα
Formula 4:
The relationship between the trigonometric function values of π - α and α can be obtained by using formula 2 and formula 3
sin（π－α）＝sinα
cos（π－α）＝－cosα
tan（π－α）＝－tanα
cot（π－α）＝－cotα
Formula 5:
The relationship between the trigonometric function values of 2 π - α and α can be obtained by using formula 1 and formula 3
sin（2π－α）＝－sinα
cos（2π－α）＝cosα
tan（2π－α）＝－tanα
cot（2π－α）＝－cotα
Formula 6:
The relationship between the trigonometric function values of π / 2 ± α and α is as follows
sin（π/2＋α）＝cosα
cos（π/2＋α）＝－sinα
tan（π/2＋α）＝－cotα
cot（π/2＋α）＝－tanα
sin（π/2－α）＝cosα
cos（π/2－α）＝sinα
tan（π/2－α）＝cotα
cot（π/2－α）＝tanα
These formulas are all for you. I don't think it will be convenient to use them
If you have to use it, please refer to the above formula
If you have to use the induction formula, I'll do it. Maybe it will be simpler
cos(60+a)+sin(30+a)
=Cos (60 + a) + cos (A-60) (this step is derived from the induction formula sin (π / 2 + α) = cos α)
=cos60cosa-sin60sina+cosacos60+sinasin60
=2cos60cosa
=cosa
Is it OK this time

Is cos (a + 60 °) = - sin (a + 30 °)?

Cos (π / 2 + α) = - sin α cos60 is equal to π / 2

If sin (30 degrees + a) = - 4 / 5, then cos (A-60 degrees) =?

cos(a-60)
=cos(60-a)
=sin[90-(60-a)]
=sin(30+a)
=-Four fifths

Simplify sin (60-a) + cos (a + 30)

sin（60-a）+cos（a+30）
=sin[90-(30+a)]+cos(a=30)
=cos(30+a) +cos(a+30)
=2cos(a+30)

Simplify sin (α + 30 °) + sin (30 ° - α) Cos α___ .

sin(α+300)+sin(300-α)
cosα=sinαcos30°+cosαsin30°+sin30°cosα-cos30°sinα
cosα
=2sin30°cosα
cosα=1，
So the answer is: 1

sin(a+30°)+cos(a+60°) 2cosa=______ .

sin(a+30°)+cos(a+60°)
2cosa=sin(a+30°)+sin(90°−a−60°)
2cosa
=sin(a+30°)+sin(30°−a)
2cosa=2cosαsin30°
2cosα=2sin30°=1，
So the answer is: 1

If the final edge of the angle α passes through the point (sin30 °, - cos30 °), sin α is equal to () A. 1 Two B. -1 Two C. - Three Two D. - Three Three

∵ if the final edge of the angle α crosses the point (sin30 °, - cos30 °),
∴x=sin30°，y=-cos30°，r=1
Then sin α = y
r=-cos30°=-
Three
Two
Therefore, C is selected

We all know that sin30 degree = 1 / 2, cos 30 degree = 2 / 2 root sign 3, then: sin square 30 degree =?, cos square 30 degree =? How to calculate the negative n power (x ^ - n) of X? For example, 2 ^ - 2

Sin squared 30 degrees = sin30 degrees * sin30 degrees = 1 / 4
Cos squared 30 degrees = cos30 degrees * cos30 degrees = 3 / 4
That is to say, sin 30 degrees, cos 30 degrees out, and then square

Simplify cos (α + 30 °) cos30 ° + sin (α + 30 °) sin30 °

cos ﹙α+30 °﹚cos30° +sin﹙α+30°﹚sin30°
=cos ﹙α+30 °-30 °﹚
=cos α