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Ymorist [56]
3 years ago
8

Derive this identity from the sum and difference formulas for cosine:

Mathematics
1 answer:
DiKsa [7]3 years ago
4 0

Step-by-step explanation:

½ [cos(a − b) − cos(a + b)]

Using sum and difference formulas:

½ [cos a cos b + sin a sin b − (cos a cos b − sin a sin b)]

Distribute the negative:

½ (cos a cos b + sin a sin b − cos a cos b + sin a sin b)

Combine like terms:

½ (2 sin a sin b)

Simplify:

sin a sin b

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85,146 rounded to the nearest thousand
larisa [96]
85,000 is your answer because The one in 146 is less then five
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Tan2(θ) − 3 = 0<br> Find all solutions for the given equation.
kondaur [170]
Hello :
tan²(θ) = 3.. equi : tan(θ) = √3  or tan(θ) = -√3
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4 0
3 years ago
Solve this pls thank you
Nana76 [90]

Answer:21

Step-by-step explanation:9+10=21

5 0
2 years ago
Please help me Solve for x. Have to show work
REY [17]
A -17 3/4< x so u have to simplify-17 3/4 by multiply 17*4=68+3=71/4 so the x stands for -71/4 and it has to be 4 as the denominator because it it the denominator for the question.
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6 0
3 years ago
Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
2 years ago
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