# How to use calculator to calculate trigonometric function?

## How to use calculator to calculate trigonometric function?

Is this not easy? Just press the "sin" key, and then press the calculated ratio, which is the degree of the angle. The other "cos" and "Tan" are also the same

### It is known that the function y = Tan ω x is at - π 2，π 2) Is a minus function, then the value range of ω is______ .

From the known condition ω < 0 and π
|ω|≥π，
∴-1≤ω＜0．
So the answer is - 1 ≤ ω < 0

### A proof of trigonometric function It is proved that 1-cosx / SiNx = SiNx / 1 + cosx

You should put a bracket. It's easy to misunderstand
（1- cosx ）/ sinx = （1- cosx ）（1+cosx）/ sinx(1+cosx)=(1-cos^2x)/sinx(1+cosx)=
sin^2x/sinx(1+cosx)=sinx/(1+cosx)
Sin ^ 2x is the square of SiNx

### A trigonometric function problem, "Oriental International School" has a triangle shaped flower bed ABC. Now it can be directly measured to ∠ a = 30 °, AC = 40m, BC = 24m. Please work out the area of this flower bed

Let the side lengths of a triangle be a, B and C respectively, so that a = 24, B = 40
Cos30 = (b * B + C * C-A * a) / (2 * b * c), you can get the value of C. then you can use s = 0.5 * b * c * Sina, and you can answer this question clearly

### Let s ω = {θ|f (x) = cos [ω (x + θ)] be an odd function}. If for each real number a, s ω∩ (a, a + 1) has no more than two elements, and there is a such that s ω∩ (a, a + 1) contains two elements, then the value range of ω is______ .

S ω = {θ|| f (x) = cos [ω (x + θ)] is an odd function} {s ω = {θ = 2K + 12 ω π, K ∈ Z} = {θ = 2K + 12 ω π, K ∈ Z} = { ，-32ωπ，-12ωπ，12ωπ，32ωπ，… }For each real number a, s ω∩ (a, a + 1) has no more than two elements, and there is a such that s ω∩ (a, a + 1) contains two elements

### Trigonometric function problem one, urgent It is known that the sides of angles a, B and C are a, B and C respectively in ＋ ABC. If a, B and C form an arithmetic sequence, B = 1, the angle a = x, a + C = f (x) 【1】 When x ∈ [(π / 6), (π / 3)], find the value range of F (x) 【2】 If f (x - π / 6) = 6 / 5, find the value of sin2x I worked out that f (x) = 2, but what's the equation for x ==

B=180°/3=60°
A+C=120°
Sine theorem:
sinx/a=sin(2π/3-x)/c=sin60°/1=√3/2
a+c=(2√3/3 ){sinx+sin(2π/3-x)}=cosx+√3sinx=2sin(x+π/6)
f(x)=2sin(x+π/6)
【1】 When x ∈ [(π / 6), (π / 3)],
The value range of F (x) [2Sin (π / 3), 2Sin (π / 2)] is [√ 3,2]
【2】 If f (x - π / 6) = 6 / 5, find the value of sin2x
f(x-π/6)=2sin(x)=5/6
sinx=5/12
cosx=√(1-25/144)=√119/12
sin2x=2sinxcosx=2*(5/12)*(√119/12)=5√119/72
)

### Find a trigonometric function problem It is known that a, B, C are the three inner angles of △ ABC, and a, B, C are the opposite sides of a, B, C respectively. The equation (x Square-1) SINB - (x-square-x) sinc - (x-1) Sina = 0 has two equal real roots. It is proved that a, B and C of the three sides are equal difference sequence, and the values of Tan (A / 2) and Tan (A / 2) are obtained

(x^2-1)sinB-(x^2-x)sinC-(x-1)sinA=0
That is (SINB sinc) x ^ 2 + (sinc Sina) x + (Sina SINB) = 0;
Δ=(sinC-sinA)^2-4(sinB-sinC)(sinA-sinB)
=(sinA)^2+4(sinB)^2+(sinC)^2-4sinAsinB+2sinAsinC-4sinBsinC
=(sinA-2sinB+sinC)^2
Because the equation (x ^ 2-1) SINB - (x ^ 2-x) sinc - (x-1) Sina = 0 has two equal real roots,
So, sin a = 0,
According to the sine theorem, there is a-2b + C = 0, so a, B, C are equal difference sequence
Tan (A / 2) can not be determined, its value range is: (0, √ 3 / 3)

### A trigonometric function problem, In the triangle ABC, it is known that sina: sinB:sinC=k : (K + 1): 2K, then the value range of K is? (the answer is half, + ∞)

According to the sine theorem: A / Sina = B / SINB = C / sinc = 2R
obtain sinA:sinB :sinC=a:b:c
A triangle requires the sum of the two sides to be greater than the third, so there is
k+k+1>2k
k+2k>k+1
k+1+2k>k
By solving the problem, we can get k > 1 / 2

### Evaluation: 2sin50 ° + sin80 ° (1+ 3tan10°) 1+cos10°．

Original formula = 2sin50 ° + 2sin80 °
cos10°(1
2cos10°+
Three
2sin10°)
2cos5°
=2sin50°+2sin80°
cos10°cos(60°−10°)
2cos5°
=2 (
Two
2sin50°+
Two
2cos50°)
cos5°
=2cos(50°−45°)
cos5°＝2．

### A problem of trigonometric function (tan20°+tan40°+tan120°)/tan20°*tan40°

For any non right triangle, there are always
tanA+tanB+tanC=tanAtanBtanC
So,
(tan20°+tan40°+tan120°)/tan20°*tan40°
=tan20°*tan40°*tan120°/tan20°*tan40°
=tan120°
=-√3
Appendix: universal formula of trigonometric function
Universal formula
(1)
(sinα)^2+(cosα)^2=1
(2)1+(tanα)^2=(secα)^2
(3)1+(cotα)^2=(cscα)^2
In order to prove the following two formulas, we only need to divide (sin α) ^ 2 and the second (COS α) ^ 2 into two parts
(4) For any non right triangle, there are always
tanA+tanB+tanC=tanAtanBtanC
Certificate:
A+B=π-C
tan(A+B)=tan(π-C)
(tanA+tanB)/(1-tanAtanB)=(tanπ-tanC)/(1+tanπtanC)
It can be sorted out
tanA+tanB+tanC=tanAtanBtanC
Get the certificate
It can also be proved that when x + y + Z = n π (n ∈ z), the relation holds
From Tana + tanb + Tanc = tanatanbtanc, the following conclusions can be drawn
(5)cotAcotB+cotAcotC+cotBcotC=1
(6)cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)
(7)(cosA）^2+(cosB）^2+(cosC）^2=1-2cosAcosBcosC
(8)（sinA）^2+（sinB）^2+（sinC）^2=2+2cosAcosBcosC
Why is trigonometric function universal formula universal
The universal formula is as follows:
Let Tan (A / 2) = t
sinA=2t/(1+t^2)
tanA=2t/(1-t^2)
cosA=(1-t^2)/(1+t^2)
That is to say sinA.tanA.cosA Tan (A / 2) can be used to express. When the maximum value of a series of functions is required, the universal formula can be used to deduce a function containing only one variable, and the maximum value is easy to find

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