How do I verify these trigonomic identities?

Question

Okay, I've been doing a review sheet for my pre-calc final and I've been able to do most of these problems okay. (Albeit it takes me awhile sometimes, lots of trial and error.)

However out of all of them these five gave me the most trouble, in the section of verifying trigonomic identities, if anyone knows how to work these out in steps I'd be greatly appreciative to know.

Anyhow here they are:

1.) sec^4 - tan^4x = sec^2x = tan^2x
2.) cos θ (sec θ - cos θ) = sin^2 θ
3.)
(sinx + cosx)^2 / sin^2x-cos^2x = sin^2x - cos^2 / (sin x - cos x)^2
(I think I'm doing this one correctly but I am still curious about it.)

4.)
(cotx - cscx)^2 = 1-cosx / 1+cosx

5.) (3cos θ - 4 sin θ)^2 + (4cos θ + 3 sin θ)^2 = 25

Any help would be great!

Sorry the correct number one is:

1.) sec^4 - tan^4x = sec^2x + tan^2x

Questions 4, and 5 are correct as typed. Any problems on that end should be taken up with my teacher who composed the review, heh.

Answer 1

I really think you'd better double-check these:

The last part of Q 1.) is CLEARLY WRONG!

sec^2 (x) = 1/cos^2 (x), and tan^2 (x) = sin^2 (x) / cos^2 (x),

so that they can ONLY be equal if sin (x) = +/- 1. In other words,
this CANNOT be an IDENTITY.

The correct result must follow from:

sec^4 x - tan^4 x = (sec^2 x - tan^2 x)(sec^2 x + tan^2 x).

But (sec^2 x - tan^2 x) = 1, so that the CORRECT identity is:

sec^4 x - tan^4 x = sec^2 x + tan^2 x.

(I suspect that you missed using the CAPS key when mistyping the '+' sign, thus obtaining an '=' sign.)

Q2.) cos θ (sec θ - cos θ) = cos θ (1/cos θ - cos θ)

= 1 - cos^2 θ = sin^2 θ.

Q's 3.) and 4.) I will tackle these when you have typed them correctly. (I suspect that you have not typed parentheses that are required to make these unambiguous mathematical expressions. As they stand, they probably violate the usual requirements for the precedence in which mathematical operations should be performed. It is up to YOU, the poser, to make your intentions clear, rather than for responders to attempt to unscramble badly posed expressions.)

Q 5.) is correct. The cross terms cancel one another, leaving

(3^2 + 4^2)(cos^2 θ + sin^2 θ) = (3^2 + 4^2) = 5^2 *** = 25.

(*** By a well-know Pythagorean identity, of course.)

Live long and prosper.

Afterword: I have now read your response to my rather mild suggestion that you typed Q's 3 and 4 incorrectly. You dismissed that possibility out of hand.

Look, you tedious squirt, I have taught mathematics in two of the world's finest universities. Yet you, who cannot even answer questions in elementary trigonometry, cannot take a broad hint that you need to examine rather carefully what you have posed as problems, because they violate standard mathematical conventions.

I very much doubt that your teacher posed these questions as YOU have represented them. (Had he/she done so, he/she should be prepared to surrender his/her teaching diploma.)

The point is that you have failed to observe the well-known rules (PEMDAS or BEMDAS, according to your source) for the order of precedence in the performance of mathematical operations.

(Note that although Helmut, below, is a tolerably decent mathematician, he too is regrettably and prolifically careless about adhering to such standard conventions. He is therefore no model to be followed.)

Q. 3. Interpreted according to the quite usual, indeed STANDARD convention(s), your Q. 3 asks us to show that:

[(sinx + cosx)^2 / sin^2x] - cos^2x

= sin^2x - [cos^2 / (sin x - cos x)^2];

analogously, your Q. 4 claims that:

(cotx - cscx)^2 = 1- [cosx / 1] + cosx = 1 (!)

WHY does your presentation mean what I have just typed? Because PEMDAS (or BEMDAS), the universally agreed order of precedence, means (i) FIRST evaluate parentheses and/or brackets, (ii) THEN evaluate exponents. (iii) THEN, perform multiplications and divisions; (iv) FINALLY, do additions and subtractions.

I hope you can now see that your quite serious CRIME AGAINST MATHEMATICAL CONVENTIONS was in failing to employ parentheses and brackets to clearly delineate the separate NUMERATORS and DENOMINATORS in your desired expressions.

Let us now re-write your Q's 3. and 4, CORRECTLY and therefore UNAMBIGUOUSLY. In order to emphasize what you failed to do, I shall employ square brackets in the initial statements of these problems, thus: ' [ ' and ' ] ', although curved parentheses ( ' ( ' and ' ) ' ) would work just as well, and would indeed be preferable as they occur at the same level of logical order:

Q. 3.) (sin x + cos x)^2 / [sin^2 x - cos^2 x]

= [sin^2 x - cos^2 [x] ] / (sin x - cos x)^2.

Here, the LHS =

(sin x + cos x)^2 / [(sin x + cos x)(sin x - cos x)]

= (sin x + cos x) / (sin x - cos x), while:

the RHS = (sin x + cos x)(sin x - cos x) / (sin x - cos x)^2

= (sin x + cos x) / (sin x - cos x).

But the latter (simplified) RHS expression is identical to the former (simplified) LHS expression. Therefore:

THE ORIGINAL TWO EXPRESSIONS ARE IDENTICALLY EQUAL.

Q.4. Correctly written this part asks us to show that:

(cot x - csc x)^2 = [1 - cos x] / [1 + cos x].

Once correctly written, this is UTTER SIMPLICITY:

(cot x - csc x)^2 = (cos x - 1)^2 / sin^2 x

= (cos x - 1)^2 / (1 - cos^2 x)

= (cos x - 1)^2 / [(1 - cos x)(1 + cos x)]

= (1 - cos x) / (1 + cos x).

This would be your finally desired conclusion, HAD it been written correctly.

As General Macarthur famously (if somewhat ambiguously) said: "These proceedings are now ended."

Answer 2

1.) sec^4 - tan^4x =
(sec^2 - tan^2x) (sec^2 + tan^2x) =
(1 + tan^2x - tan^2x) (sec^2 + tan^2x) =
sec^2 + tan^2x
2.) cos θ (sec θ - cos θ) =? sin^2 θ
cos θ/cos θ - cos^2 θ =? sin^2 θ
1 - cos^2 θ = sin^2 θ
sin^2 θ + cos^2 θ - cos^2 θ = sin^2 θ
sin^2 θ = sin^2 θ
3.)
(sinx + cosx)^2 / sin^2x-cos^2x = sin^2x - cos^2 / (sin x - cos x)^2
(sinx + cosx)^2 / (sinx + cosx)(sinx - cosx) = sin^2x - cos^2 / (sin x - cos x)^2
(sinx + cosx) / (sinx - cosx) = sin^2x - cos^2 / (sin x - cos x)^2
(sinx + cosx)(sinx - cosx) / (sinx - cosx)^2 = sin^2x - cos^2 / (sin x - cos x)^2
(sin^2x - cos^2x) / (sinx - cosx)^2 = sin^2x - cos^2 / (sin x - cos x)^2
4.)
(cotx - cscx)^2 = 1-cosx / 1+cosx
(cot^2x - 2cotxcscx + csc^2x) = 1-cosx / 1+cosx
(1 + 2cos^2x/sin^2x - 2cosx/sin^2x ) = 1-cosx / 1+cosx
((sin^2x + 2cos^2x - 2cosx)/sin^2x ) = 1-cosx / 1+cosx
((1 + cos^2x - 2cosx)/(1 - cos^2x ) = 1-cosx / 1+cosx
((1 - cosx)^2/(1 - cosx)( 1 + cosx) = 1-cosx / 1+cosx
(1 - cosx)/( 1 + cosx) = 1-cosx / 1+cosx
5.)
(3cos θ - 4 sin θ)^2 + (4cos θ + 3 sin θ)^2 = 25
9cos^2θ - 24sinθcosθ + 16sin^2θ + 16cos^2θ + 24sinθcosθ + 9sin^2θ = 25
9cos^2θ + 9sin^2θ + 16sin^2θ + 16cos^2θ = 25
9(cos^2θ + sin^2θ) + 16(sin^2θ + cos^2θ) = 25
9 + 16 = 25

Answer 3

LHS= cos(x+y)/cosxsiny =[cosxcosy - sinxsiny]/cosxsiny = cosxcosy/cosxsiny - sinxsiny/cosxsiny = cosy/siny - sinx/cosx = coty - tanx Hmmmm