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Maksim231197 [3]
3 years ago
11

Determine whether the statements in (a) and (b) are logically equivalent.

Mathematics
2 answers:
sertanlavr [38]3 years ago
8 0

Answer:

Not 100% but I think they are equal logically.

Andru [333]3 years ago
5 0

Answer:

The two statements are logically equivalent.

Step-by-step explanation:

Let X be the statement that Bob is a double math and computer science major.

Y be the statement that Ann is a maths major.

Z be the statement that Ann is a double maths and computer science major.

The two statements written in terms of X, Y and Z now

a. Bob is a double math and computer science major and Ann is a math major, but Ann is not a double math and computer science major.

(x and y) and not z

b. It is not the case that both Bob and Ann are double math and computer science majors, but it is the case that Ann is a math major and Bob is a double math and computer science major.

not (x and z) and (x and y)

Noting that for logical statements,

Negation is represented by ~

And is represented by conjunction sign Λ

Or is represented by disjunction sign V

(x and y) and not z

(x Λ y) Λ (~z)

not (x and z) and (x and y)

~(x Λ z) Λ (x Λ y)

We can then simplify the second statement to obtain the first statement and prove the equivalence of both sides

~(x Λ z) Λ (x Λ y)

Using DE MORGAN'S theory, ~(x Λ z) = (~x) V (~z)

~(x Λ z) Λ (x Λ y) = ((~x) V (~z)) Λ (x Λ y)

Then applying the distributive law to the expression, we can open the bracket up

((~x) V (~z)) Λ (x Λ y)

= ((~x) Λ (x Λ y)) V ((~z) Λ (x Λ y))

Opening the first bracket up further

((~x) Λ (x Λ y)) V ((~z) Λ (x Λ y))

= ((~x) Λ x) Λ y) V ((~z) Λ (x Λ y))

The NEGATION law shows that (~x Λ x) = c (where c is a negation law parameter for when two opposite statements are combined in this manner, it works like a 0 in operation)

((~x) Λ x) Λ y) V ((~z) Λ (x Λ y))

= (c Λ y) V ((~z) Λ (x Λ y))

But (c Λ y) = (y Λ c) = c (according to the UNIVERSALLY BOUND law, see how c works like a 0 now?)

(c Λ y) V ((~z) Λ (x Λ y))

= c V ((~z) Λ (x Λ y))

= ((~z) Λ (x Λ y)) V c (commutative)

And one of the foremost IDENTITY laws is that (any statement) V c = c

((~z) Λ (x Λ y)) V c

= ((~z) Λ (x Λ y))

= (x Λ y) Λ (~z)

Which is the same as the first statement!

PROVED!!!

Hope this Helps!!!

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cos2t/cos²t

Step-by-step explanation:

Here the given trigonometric expression to us is ,

\longrightarrow \dfrac{cos^4t - sin^4t }{cos^2t }

We can write the numerator as ,

\longrightarrow \dfrac{ (cos^2t)^2-(sin^2t)^2}{cos^2t }

Recall the identity ,

\longrightarrow (a-b)(a+b)=a^2-b^2

Using this we have ,

\longrightarrow \dfrac{(cos^2t + sin^2t)(cos^2t-sin^2t)}{cos^2t}

Again , as we know that ,

\longrightarrow sin^2\phi + cos^2\phi = 1

Therefore we can rewrite it as ,

\longrightarrow \dfrac{1(cos^2t - sin^2t)}{cos^2t}

Again using the first identity mentioned above ,

\longrightarrow \underline{\underline{\dfrac{(cost + sint )(cost - sint)}{cos^2t}}}

Or else we can also write it using ,

\longrightarrow cos2\phi = cos^2\phi - sin^2\phi

Therefore ,

\longrightarrow \underline{\underline{\dfrac{cos2t}{cos^2t}}}

And we are done !

\rule{200}{4}

Additional info :-

<em>D</em><em>e</em><em>r</em><em>i</em><em>v</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>o</em><em>f</em><em> </em><em>c</em><em>o</em><em>s</em><em>²</em><em>x</em><em> </em><em>-</em><em> </em><em>s</em><em>i</em><em>n</em><em>²</em><em>x</em><em> </em><em>=</em><em> </em><em>c</em><em>o</em><em>s</em><em>2</em><em>x</em><em> </em><em>:</em><em>-</em>

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As we know that ,

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So that ,

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On simplifying,

\longrightarrow cos(x+x) = cos^2x - sin^2x

Hence,

\longrightarrow\underline{\underline{cos (2x) = cos^2x - sin^2x }}

\rule{200}{4}

7 0
2 years ago
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