Math, asked by praveenkrpandey8960, 4 months ago

cos3x बराबर क्या होता है​

Answers

Answered by rushikeshphapale4
1

Answer:

cos(3x) = 4cos^3(x) − 3cos(x)

Here are some Important results.

Pythagorean Identities

sin 2X + cos 2X = 1

1 + tan 2X = sec 2X

1 + cot 2X = csc 2X

Negative Angle Identities

sin (-X) = – sin X, odd function

csc (-X) = – csc X, odd function

cos (-X) = cos X, even function

sec (-X) = sec X, even function

tan (-X) = – tan X, odd function

cot (-X) = – cot X, odd function

Cofunctions Identities

sin (π /2 – X) = cos X

cos (π /2 – X) = sin X

tan (π /2 – X) = cot X

cot (π/2 – X) = tan X

sec (π /2 – X) = csc X

csc (π /2 – X) = sec X

Addition Formulas

cos (X + Y) = cos X cos Y – sin X sin Y

cos (X – Y) = cos X cos Y + sin X sin Y

sin (X + Y) = sin X cos Y + cos X sin Y

sin (X – Y) = sin X cosY – cos X sin Y

tan (X + Y) = [ tan X + tan Y ] / [ 1 – tan X tan Y]

tan (X – Y) = [ tan X – tan Y ] / [ 1 + tan X tan Y]

cot (X + Y) = [ cot X cot Y – 1 ] / [ cot X + cot Y]

cot (X – Y) = [ cot X cot Y + 1 ] / [ cot Y – cot X]

Sum to Product Formulas

cos X + cos Y = 2cos[(X + Y)/ 2] cos[(X – Y)/ 2]

sin X + sin Y = 2sin[(X + Y)/ 2] cos[(X – Y)/ 2]

Difference to Product Formulas

cos X – cos Y = – 2sin[(X + Y) / 2] sin[(X – Y) / 2]

sin X – sin Y = 2cos[(X + Y) / 2] sin[(X – Y) / 2]

Product to Sum/Difference Formulas

cos X cos Y = (1/2) [cos (X – Y) + cos (X + Y)]

sin X cos Y = (1/2) [sin (X + Y) + sin (X – Y)]

cos X sin Y = (1/2) [sin (X + Y) – sin[ (X – Y)]

sin X sin Y = (1/2) [cos (X – Y) – cos (X + Y)]

Difference of Squares Formulas

sin 2X – sin 2Y = sin (X + Y) sin (X – Y)

cos 2X – cos 2Y = – sin (X + Y) sin (X – Y)

cos 2X – sin 2Y = cos (X + Y) cos (X – Y)

Double Angle Formulas

sin (2X) = 2 sin X cos X

cos (2X) = 1 – 2sin 2X = 2cos 2X – 1

tan (2X) = 2tan X/[1 – tan 2X]

Multiple Angle Formulas

sin (3X) = 3sin X – 4sin 3X

cos (3X) = 4cos 3X – 3cos X

sin (4X) = 4sin X cos X – 8sin 3X cos X

cos (4X) = 8cos 4X – 8cos 2X + 1

Half Angle Formulas

sin (X/2) = ±√[(1 – cos X)/2]

cos (X/2) = ±√[(1 + cos X)/2]

tan (X/2) = ±√[(1 – cos X)/(1 + cos X)]

= sin X/(1 + cos X)

= (1 – cos X)/sin X

Power Reducing Formulas

sin 2X = 1/2 – (1/2) cos (2X))

cos 2X = 1/2 + (1/2) cos (2X))

sin 3X = (3/4) sin X – (1/4) sin (3X)

cos 3X = (3/4) cos X + (1/4) cos (3X)

sin 4X = (3/8) – (1/2)cos (2X) + (1/8)cos (4X)

cos 4X = (3/8) + (1/2)cos (2X) + (1/8)cos (4X)

sin 5X = (5/8)sin X – (5/16)sin (3X) + (1/16)sin (5X)

cos 5X = (5/8)cos X + (5/16)cos (3X) + (1/16)cos (5X)

sin 6X = 5/16 – (15/32)cos (2X) + (6/32)cos (4X) – (1/32)cos (6X)

cos 6X = 5/16 + (15/32)cos (2X) + (6/32)cos (4X) + (1/32)cos (6X)

·

Trigonometric Functions Periodicity

sin (X + 2π) = sin X, period 2π

cos (X + 2π) = cos X, period 2π

sec (X + 2π) = sec X, period 2π

csc (X + 2π) = csc X, period 2π

tan (X + π) = tan X, period π

cot (X + π) = cot X, period π

Hope this will help

Answered by vinayakdev959
4

Step-by-step explanation:

ANSWER

What is cos3x equal to? How can you prove it?

=Cos(3x)=4Cos3(x)−3Cos(x)

Cos(3x)=Cos(2x+x)=Cos(2x)Cos(x)−Sin(2x)Sin(x)

=[2Cos2(x)−1]Cos(x)−2Sin(x)Cos(x)Sin(x)

=2Cos3(x)−Cos(x)−2Cos(x)[Sin2(x)]

=2Cos3(x)−Cos(x)−2Cos(x)[1−Cos2(x)]

=2Cos3(x)−Cos(x)−2Cos(x)+2Cos3(x)

=4Cos3(x)−3Cos(x)

See, I don’t know how to use the [maths] stuff..

I tried to write it the best way.

Basically, put 3x=2x+x and use the identity Cos(A+B)

After that, using other identities of Cos(2A) and Sin(2A) , convert all terms into cos. Open the brackets, add, subtract, and voilà, you got it!!

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