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sin30°＋cot45°＋tan60°－cot30°
=1/2+1+√3-√3
=3/2

What is the meaning of COS 45, sin 30 and COS 60 in mathematical problems?

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A = 270 ° or a = 30 ° or 1 / 2 + Sina = root 3 / 2 * cosa + 1 / 2 * Sina = root 3 / 2 * cosa-1 / 2 Sina = root 3 * cosa-1 / 2 Sina = root 3 * cosa-1 substituting (Sina) ^ 2 + (COSA) ^ 2 = 1 cosa (2cosa root 3) = 0, cosa = 0 or cosa = 3 / 2 ﹣ a = 3 π / 2 + 2K

Let f (Sina + COSA) = Sina * cosa, then the value of F (sin 30 °) My approach is, Let's set an angle X sinx+cosx=sin30° √ 2(√ 2/2*sinx+√ 2/2*cosx)=sin30° √ 2sin(45°+x)=sin30° Sin (45 ° + x) = √ 2 / 2 = sin45 ° or sin135 ° Then x is 0 ° or 90 ° Then f (sin30 °) = f (sin0 ° + cos0 °) = 0 Or F (sin90 ° + cos90 °) = 0 The answer is - 3 / 8

 
sinx+cosx=sin30°
√2(√2/2*sinx+√2/2*cosx)=sin30°
√2sin(45°+x)=sin30°{√2sin(45°+x)=sin30°=1/2}
Error × sin (45 ° + x) = √ 2 / 2 = sin 45 ° or sin 135 °
sin(45°+x)=1/2
x=-15°

Let f (Sina + COSA) = Sina * cosa, then the value of F (sin30 degree) is obtained

F (Sina + COSA) = sinacosa = (2sinacosa + 1) / 2-1 / 2 = the square of (Sina + COSA) / 2-1 / 2, that is, f (x) = the square of X / 2-1 / 2
So f (sin30) = 1 / 8-1 / 2 = - 3 / 8

If f (x) = Sina cosa, then f '(a) is equal to

If a is a constant
Then f (x) = Sina cosa is a constant function
So f '(x) = 0
So '(a) = 0

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f(sina)=f(cos(π/2-a))
=cos[17(π/2-a)]
=cos(17π/2-17a)
=cos[8π+(π/2-17a)]
=cos(π/2-17a)
=sin17a

F (Sina + COSA) = sinacosa f (COS 30 degrees)=

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To solve an equation with trigonometric function, please explain the solution (teach me how to solve it) I'm on the third day of junior high school. Please be more specific Take the following question as an example: sinC/8=Sin（3∠C）/10

sin(3∠C)=3sinC-4(sinC)^3
Replace it and solve the equation

Solution of trigonometric function equation There are two equations, Mg, q for F and Q for a; Fcos(Q-a)=4mgsinQcosQ; F/sin(2Q)=mg/cos(a+Q); The specific derivation process is found; I just graduated from junior high school, and have a little understanding of trigonometric function; There is a need for questions,

If yes, use mg, q for F and Q for a;
Fcos(Q-a)=4mgsinQcosQ;
F/sin(2Q)=mg/cos(a+Q);
The words of
I think this is beyond the scope of junior high school mathematics knowledge. To use trigonometric identity transformation belongs to the knowledge of high school mathematics. I don't know if you can't understand it. I'll calculate it first
cos（Q-a）=cosQcosa+sinQsina sin（2Q）=2sinQcosQ
cos（a+Q）=cosacosQ-sinasinQ
Then: fcos (Q-A) = 4mgsinqcosq; F = 4mgsinqcosq / cos (Q-A)
=4mgsinQcosQ/（cosQcosa+sinQsina）
From F / sin (2q) = mg / cos (a + Q), f = 2mgsinqcosq / (cosacos q-sinasin q)
That is, 4mgsinqcosq / (cosqcosa + sinqsina) = 2mgsinqcosq / (cosacosq sinasin q)
2 =（cosQcosa+sinQsina）/（cosacosQ-sinasinQ）
=cosQcosa-3sinQsina =0
Cosq cosa = 3sinqsina: cosq / 3sinq = cosa / Sina: cotq = 3cota
If in the first quadrant, then a = 60 ° q = 30 ° satisfies the condition, you can also convert it into radians: π radian = 180 °
If the equation holds no matter what value a is taken, it is: q = (A-30 °) + k * 180 ° {K can be taken as 0,1,2,3.. n}
Use mg, q for F, and you can deduce it by yourself
I don't think I should waste your and my time because it is far beyond the knowledge of junior high school. I'm afraid you can't understand
There may be errors in the above calculation process. I think your question is not detailed enough
Fcos(Q-a)=4mgsinQcosQ;
F / sin (2q) = mg / cos (a + Q); the derived result is still known about these two equations to deduce: Mg, q for F, q for a? Now it's too late to ask me any questions
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