1 to high school math problem proof triangle ABC, certificate cos cos cos C 1 2 cos cos cosC

Observing the original formula, we can assume that cos = 0, and there is (cos) 2+(cos) 2=1, then + = 2, and then = 2, then + = can make the original formula true if the range of values of , is not limited;

Then verify that when , 0, 2), += can make the original formula hold: since the three angles are less than 2, and the original formula must be satisfied, the sum of the three angles can only be one case; And when it is verified that 0, 2), += still makes the original formula hold.

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2b=a+c is derived from the sinusoidal theorem: 2sinb=sina+sinc utilizes the sum product formula: sina + sinc=2sin[(a+c) 2]cos[(a-c) 2] 2sin(b 2)cos(b 2) =sin[(a+c) 2]cos[(a-c) 2] In the triangle ABC, a+ b+ c= b 2 = a+c) 2 is the acute angle 2sin(b 2....

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Senior 1 math problem: In the triangle ABC, it is known that A=2, B=1, if B has two solutions, find the COSC value range...

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b is worth it, why are there still two solutions?

a=2,b=1,c=2a。Seek c

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With the cosine theorem: right = b (a + b -c ) 2ab + c (a + c -b ) 2ac

a²+b²-c²)/2a+(a²+c²-b²)/2a=2a²/2a

a=left.

Do it yourself, just talk about the method, first draw a picture, consider two situations, obtuse angle and acute angle triangle, through A, B two points respectively to do AB perpendicular line, you can find and respectively Ca, CB on the line segment, and then you know, right?

Right triangle, b2+c2=a2

The cosa is expressed by the cosine formula, which is simplified to obtain the above formula.

1+cosa=b/c+1

cosa=b c=sinb sinc (sinusoidal theorem) cosasinc=sinb

cosasinc=sin(a+c)=sinacosc+cosasinc

sinacosc=0

Since sina >0, cosc=0 and c is a right angle.

The triangle is at right angles

Use the cosine theorem to simplify,Convert cosa to ABC representation.。。 Then simplify, not surprisingly, this kind of problem is either isosceles, equilateral, or right angle You try it.